I want to prove this relation which I know is true for even n's $${{{\rm 2}^{{\it n}}\cdot{\rm (}{\it n}{\rm !)}^{{\rm 2}}}\over{{\rm (}{\rm 2}{\it n}{\rm )!}}}\times{{{\rm (}-{\rm 1}{\rm )}^{{{{\it n}}\over{{\rm 2}}}}\times\left[{{\rm 2}\times{\rm 4}\times\cdots\times{\it n}}\right]\times\left[{{\rm (}{\it n}+{\rm 1}{\rm )(}{\it n}+{\rm 3}{\rm )}\cdots\times{\rm (}{\rm 2}{\it n}-{\rm 1}{\rm )}}\right]}\over{{\it n}{\rm !}}}={\rm 1}+\sum\limits_{{\it k}{\rm =1}}^{{{{\it n}}\over{{\rm 2}}}}{{{{\rm (}-{\rm 1}{\rm )}^{{\it k}}\times\left[{\left({{\it n}-{\rm (}{\rm 2}{\it k}-{\rm 2}{\rm )}}\right)\cdots{\rm (}{\it n}-{\rm 4}{\rm )(}{\it n}-{\rm 2}{\rm )}{\it n}}\right]\left[{{\rm (}{\it n}+{\rm 1}{\rm )(}{\it n}+{\rm 3}{\rm )}\cdots\left({{\it n}+{\rm (}{\rm 2}{\it k}-{\rm 1}{\rm )}}\right)}\right]}\over{{\rm (}{\rm 2}{\it k}{\rm )!}}}}$$ I came across this formula in a textbook. I know this is true for all even positive integers n. For example, for n=4, the left side of equality is: $${{2^{4}\cdot (4!)^{2}}\over{8!}}\times{{(-1)^{2}\times (2\times 4)\times (5\times 7)}\over{4!}}={{8}\over{3}}$$ and the right side is: $$1+\sum\limits_{k=1}^{2}{{{(-1)^{k}\left[{\left({4-(2k-2)}\right)\times\cdots\times (4-2)\times 4}\right]\times\left[{(4+1)(4+3)\cdots (4+(2k-1))}\right]}\over{(2k)!}}}=1+{{(-1)\times 4\times 5}\over{2!}}+{{(-1)^{2}\times 2\times 4\times 5\times 7}\over{4!}}=1-10+{{35}\over{3}}{{35-27}\over{3}}={{8}\over{3}}$$ as you can see it's true for n=4.
Edit: I used induction to prove this formula. As said before, we know that this relation is true for even n's, so first we show that the relationship is true for n=2. for n=2 the left side is $${{2^{2}\times (2!)^{2}}\over{4!}}\times{{(-1)\times 2\times 3}\over{2!}}=-2$$ and the right side is $$1+{{(-1)\times 2\times 3}\over{2!}}=1-3=-2$$ so it's true for n=2. Now, we assume that the relation is true for an arbitrary even number n, that is $${{{\rm 2}^{{\it n}}\cdot{\rm (}{\it n}{\rm !)}^{{\rm 2}}}\over{{\rm (}{\rm 2}{\it n}{\rm )!}}}\times{{{\rm (}-{\rm 1}{\rm )}^{{{{\it n}}\over{{\rm 2}}}}\times\left[{{\rm 2}\times{\rm 4}\times\cdots\times{\it n}}\right]\times\left[{{\rm (}{\it n}+{\rm 1}{\rm )(}{\it n}+{\rm 3}{\rm )}\cdots\times{\rm (}{\rm 2}{\it n}-{\rm 1}{\rm )}}\right]}\over{{\it n}{\rm !}}}={\rm 1}+\sum\limits_{{\it k}{\rm =1}}^{{{{\it n}}\over{{\rm 2}}}}{{{{\rm (}-{\rm 1}{\rm )}^{{\it k}}\times\left[{\left({{\it n}-{\rm (}{\rm 2}{\it k}-{\rm 2}{\rm )}}\right)\cdots{\rm (}{\it n}-{\rm 4}{\rm )(}{\it n}-{\rm 2}{\rm )}{\it n}}\right]\left[{{\rm (}{\it n}+{\rm 1}{\rm )(}{\it n}+{\rm 3}{\rm )}\cdots\left({{\it n}+{\rm (}{\rm 2}{\it k}-{\rm 1}{\rm )}}\right)}\right]}\over{{\rm (}{\rm 2}{\it k}{\rm )!}}}}$$ And we try to show that the relation is also true for n+2, that is, to prove that $${{{\rm 2}^{{\rm n}{\rm +}{\rm 2}}{\rm \cdot}{\rm ((n}{\rm +}{\rm 2)!)}^{{\rm 2}}}\over{{\rm (2(n}{\rm +}{\rm 2))!}}}{\rm \times}{{{\rm (}{\rm -}{\rm 1)}^{{{{\rm n}{\rm +}{\rm 2}}\over{{\rm 2}}}}{\rm \times}\left[{{\rm 2}{\rm \times}{\rm 4}{\rm \times}{\rm \cdots}{\rm \times}{\rm (n}{\rm +}{\rm 2)}}\right]{\rm \times}\left[{{\rm (n}{\rm +}{\rm 3}{\rm )(n}{\rm +}{\rm 5}{\rm )}{\rm \cdots}{\rm \times}{\rm (2n}{\rm +}{\rm 3)}}\right]}\over{{\rm (}{\rm n}{\rm +}{\rm 2)}{\rm !}}}{\rm =}{\rm 1}{\rm +}\sum\limits_{{\rm k}{\rm =}{\rm 1}}^{{{{\rm n}{\rm +}{\rm 2}}\over{{\rm 2}}}}{{{{\rm (}{\rm -}{\rm 1)}^{{\rm k}}{\rm \times}\left[{\left({{\rm n}{\rm +}{\rm 2}{\rm -}{\rm (2k}{\rm -2}{\rm )}}\right){\rm \cdots}{\rm (n}{\rm -}{\rm 4)n(n}{\rm +}{\rm 2)}}\right]\left[{{\rm (n}{\rm +}{\rm 3)(n}{\rm +}{\rm 5}{\rm )}{\rm \cdots}\left({{\rm n}{\rm +}{\rm 2}{\rm +}{\rm (2k}{\rm -}{\rm 1}{\rm )}}\right)}\right]}\over{{\rm (2k)!}}}}$$ or $${{{\rm 2}^{{\rm n}{\rm +}{\rm 2}}{\rm \cdot}{\rm ((n}{\rm +}{\rm 2)!)}^{{\rm 2}}}\over{{\rm (2(n}{\rm +}{\rm 2))!}}}{\rm \times}{{{\rm (}{\rm -}{\rm 1)}^{{{{\rm n}{\rm +}{\rm 2}}\over{{\rm 2}}}}{\rm \times}\left[{{\rm 2}{\rm \times}{\rm 4}{\rm \times}{\rm \cdots}{\rm \times}{\rm (n}{\rm +}{\rm 2)}}\right]{\rm \times}\left[{{\rm (n}{\rm +}{\rm 3}{\rm )(n}{\rm +}{\rm 5}{\rm )}{\rm \cdots}{\rm \times}{\rm (2n}{\rm +}{\rm 3)}}\right]}\over{{\rm (}{\rm n}{\rm +}{\rm 2)}{\rm !}}}{\rm =}{\rm 1}{\rm +}\sum\limits_{{\rm k}{\rm =}{\rm 1}}^{{{{\rm n}{\rm +}{\rm 2}}\over{{\rm 2}}}}{{{{\rm (}{\rm -}{\rm 1)}^{{\rm k}}{\rm \times}\left[{\left({{\rm n}{\rm -}{\rm (2k}{\rm -4}{\rm )}}\right){\rm \cdots}{\rm (n}{\rm -}{\rm 4)n(n}{\rm +}{\rm 2)}}\right]\left[{{\rm (n}{\rm +}{\rm 3)(n}{\rm +}{\rm 5}{\rm )}{\rm \cdots}\left({{\rm n}{\rm +}{\rm (2k}{\rm +}{\rm 1)}}\right)}\right]}\over{{\rm (2k)!}}}}$$ Second Edit: Using the relation ${{{\rm 2}^{{\rm n}}{\rm \cdot}{\left({{\rm n!}}\right)}^{{\rm 2}}}\over{{\rm (2n)!n!}}}{\rm =}{{{\rm 1}}\over{{\rm (2n-1)!!}}}$ that FShrike suggested, the desired relation is rewritten in a simpler form as follows: $${{{\rm (}{\rm -}{\rm 1}{\rm )}^{{{{\rm n}}\over{{\rm 2}}}}{\rm \times}\left[{{\rm 2}{\rm \times}{\rm 4}{\rm \times}{\rm \cdots}{\rm \times}{\rm n}}\right]}\over{{\rm 1*3*...*(n-3)*(n-1)}}}{\rm =1+}\sum\limits_{{\rm k=1}}^{{{{\rm n}}\over{{\rm 2}}}}{{{{\rm (}{\rm -}{\rm 1}{\rm )}^{{\rm k}}{\rm \times}\left[{\left({{\rm n}{\rm -}{\rm (}{\rm 2k}{\rm -}{\rm 2}{\rm )}}\right){\rm \cdots}{\rm (}{\rm n}{\rm -}{\rm 2}{\rm )n}}\right]\left[{{\rm (n}{\rm +}{\rm 1)(}{\rm n}{\rm +}{\rm 3}{\rm )}{\rm \cdots}\left({{\rm n}{\rm +}{\rm (}{\rm 2k}{\rm -}{\rm 1}{\rm )}}\right)}\right]}\over{{\rm (}{\rm 2k}{\rm )!}}}}$$ and we want to conclude the following relationship from it: $${{{\rm (}{\rm -}{\rm 1)}^{{{{\rm n}}\over{{\rm 2}}}{\rm +1}}{\rm \times}\left[{{\rm 2}{\rm \times}{\rm 4}{\rm \times}{\rm \cdots}{\rm \times}{\rm (n}{\rm +}{\rm 2)}}\right]}\over{{\rm 1*3*...*(n}{\rm +}{\rm 1)}}}{\rm =1+}\sum\limits_{{\rm k}{\rm =}{\rm 1}}^{{{{\rm n+2}}\over{{\rm 2}}}}{{{{\rm (}{\rm -}{\rm 1)}^{{\rm k}}{\rm \times}\left[{\left({{\rm n}{\rm -}{\rm (2k}{\rm -4}{\rm )}}\right){\rm \cdots}{\rm (n}{\rm -2}{\rm )n(n}{\rm +}{\rm 2)}}\right]\left[{{\rm (n}{\rm +}{\rm 3)(n}{\rm +}{\rm 5}{\rm )}{\rm \cdots}\left({{\rm n}{\rm +}{\rm (2k}{\rm +}{\rm 1}{\rm )}}\right)}\right]}\over{{\rm (2k)!}}}}$$ By comparing the sides of the last two relations, we come to the conclusion that we must multiply the sides of the first relation by ${\rm -}{{{\rm n+2}}\over{{\rm n+1}}}$. This yields: $${{{\rm (}{\rm -}{\rm 1)}^{{{{\rm n}}\over{{\rm 2}}}{\rm +1}}{\rm \times}\left[{{\rm 2}{\rm \times}{\rm 4}{\rm \times}{\rm \cdots}{\rm \times}{\rm (n}{\rm +}{\rm 2)}}\right]}\over{{\rm 1*3*...*(n}{\rm +}{\rm 1)}}}{\rm =-}{{{\rm n+2}}\over{{\rm n+1}}}{\rm +}\sum\limits_{{\rm k}{\rm =}{\rm 1}}^{{{{\rm n}}\over{{\rm 2}}}}{{{{\rm (}{\rm -}{\rm 1)}^{{\rm k}{\rm +}{\rm 1}}{\rm \times}\left[{\left({{\rm n}{\rm -}{\rm (2k}{\rm -2}{\rm )}}\right){\rm \cdots}{\rm (n}{\rm -2}{\rm )n(n}{\rm +}{\rm 2)}}\right]\left[{{\rm (n}{\rm +}{\rm 3)(n}{\rm +}{\rm 5}{\rm )}{\rm \cdots}\left({{\rm n}{\rm +}{\rm (2k}{\rm -}{\rm 1}{\rm )}}\right)}\right]}\over{{\rm (2k)!}}}}$$ The left side of the relationship is correct. Now we have to change the right side so that it fits into the desired form. So now we must prove $${\rm 1+}\sum\limits_{{\rm k}{\rm =}{\rm 1}}^{{{{\rm n+2}}\over{{\rm 2}}}}{{{{\rm (}{\rm -}{\rm 1)}^{{\rm k}}{\rm \times}\left[{\left({{\rm n}{\rm -}{\rm (2k}{\rm -4}{\rm )}}\right){\rm \cdots}{\rm (n}{\rm -2}{\rm )n(n}{\rm +}{\rm 2)}}\right]\left[{{\rm (n}{\rm +}{\rm 3)(n}{\rm +}{\rm 5}{\rm )}{\rm \cdots}\left({{\rm n}{\rm +}{\rm (2k}{\rm +}{\rm 1}{\rm )}}\right)}\right]}\over{{\rm (2k)!}}}}$$ $${\rm =-}{{{\rm n+2}}\over{{\rm n+1}}}{\rm +}\sum\limits_{{\rm k}{\rm =}{\rm 1}}^{{{{\rm n}}\over{{\rm 2}}}}{{{{\rm (}{\rm -}{\rm 1)}^{{\rm k}{\rm +}{\rm 1}}{\rm \times}\left[{\left({{\rm n}{\rm -}{\rm (2k}{\rm -2}{\rm )}}\right){\rm \cdots}{\rm (n}{\rm -2}{\rm )n(n}{\rm +}{\rm 2)}}\right]\left[{{\rm (n}{\rm +}{\rm 3)(n}{\rm +}{\rm 5}{\rm )}{\rm \cdots}\left({{\rm n}{\rm +}{\rm (2k}{\rm -}{\rm 1}{\rm )}}\right)}\right]}\over{{\rm (2k)!}}}}$$
Hint: Often it is convenient to try to simplify identities at first. Since $n$ is assumed to be an even non-negative integer we will take $n=2m$ in the following. We show OPs claimed identity is equivalent to show \begin{align*} \color{blue}{\sum_{k=0}^m(-1)^k\binom{4m-2k}{2m}\binom{2m}{k}=4^m\qquad\qquad m\geq 0}\tag{1} \end{align*} which might also be an interesting starting point to prove the identity.
In order to do the transformations we use factorials, double-factorials, binomial coefficients and a relationship between even and odd double factorials. \begin{align*} m!&=m\cdot(m-1)\cdots3\cdot2\cdot 1\tag{2.1}\\ (2m)!!&=(2m)(2m-2)\cdots 6\cdot4\cdot2=2^m\cdot m!\tag{2.2}\\ (2m-1)!!&=(2m-1)(2m-3)\cdots 5\cdot 3\cdot 1=\frac{(2m)!}{(2m)!!}=\frac{(2m)!}{2^mm!}\tag{2.3}\\ \binom{m}{k}&=\frac{m!}{k!(m-k)!}\tag{2.4} \end{align*}
We start with OPs left-hand side and obtain \begin{align*} &\frac{2^{2m}(2m)!(2m)!}{(4m)!}(-1)^m\frac{(2m)!!}{(2m)!}\,\frac{(4m-1)!!}{(2m-1)!!}\tag{$\to\ (2.2),(2.3)$}\\ &\qquad=\frac{2^{2m}(2m)!(2m)!}{(4m)!}(-1)^m\frac{2^mm!}{(2m)!}\,\frac{(4m)!}{2^{2m}(2m)!}\,\frac{2^mm!}{(2m)!}\tag{$\to\ (2.3)$}\\ &\qquad=(-1)^m2^{2m}\frac{(m!)^2}{(2m)!}\tag{$\to\ $cancellation}\\ &\qquad\,\,\color{blue}{=(-1)^m4^m\binom{2m}{m}^{-1}}\tag{$\to\ (2.4)$} \end{align*}
Simplification of OPs right-hand side gives \begin{align*} &1+\sum_{k=1}^m\frac{(-1)^k}{(2k)!}\,\frac{(2m)!!}{(2m-2k)!!}\,\frac{(2m+2k-1)!!}{(2m-1)!!}\tag{$\to\ (2.2),(2.3)$}\\ &\quad=1+\sum_{k=1}^m\frac{(-1)^k}{(2k)!}\,\frac{2^mm!}{2^{m-k}(m-k)!}\,\frac{(2m+2k)!}{2^{m+k}(m+k)!}\,\frac{2^mm!}{(2m)!}\tag{$\to\ (2.2),(2.3)$}\\ &\quad=1+\binom{2m}{m}^{-1}\sum_{k=1}^m\frac{(-1)^k}{(2k)!}\,\frac{(2m+2k)!}{(m-k)!(m+k)!}\tag{$\to\ (2.4)$, cancellation}\\ &\quad\,\,\color{blue}{=1+\binom{2m}{m}^{-1}\sum_{k=1}^m(-1)^k\binom{2m+2k}{2k}\binom{2m}{m+k}}\tag{$\to \ (2.4)$} \end{align*}
Comment:
In (4.1) we multiply with $\binom{2m}{m}$.
In (4.2) we can merge $1$ into the sum by starting with $k=0$.
in (4.3) we change the order of summation $k\to m-k$.
In (4.4) we use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$ twice and multiply with $(-1)^m$.