Prove that $$ a_n\bigg[1-\frac{n(n-1)}{2(2n-1)}+\frac{n(n-1)(n-2)(n-3)}{2\cdot4\cdot(2n-1)(2n-3)}-\cdots+\frac{n(n-1)(n-2)\cdots(n-2k+1)}{2\cdot4\cdots 2k\cdot(2n-1)(2n-3)\cdots(2n-2k+1)}\bigg]=1\\ \implies a_n=\frac{1}{1+\sum_{k=1}^{p} \frac{(-1)^k.n(n-1)(n-2)\cdots(n-2k+1)}{2^k\cdot k!\cdot(2n-1)(2n-3)\cdots(2n-2k+1)}}=\frac{(2n-1)(2n-3)\cdots 3\cdot1}{n!}=\frac{(2n)!}{2^n(n!)^2} $$ where $p=n/2$ if $n$ is even.
This actually comes in the derivation of the Legendre's polynomials as the solutions of the Legendre's differential equation, when we try to do the standardization.
We can verify it for each $n$ but is it possible to directly evaluate the series to the required form ?
First let's rewrite everything using factorials. Since $1\cdot 3\cdots(2n-1)=(2n)!/(2^n n!)$ (including $n=0$ if we interpret the "product of zero terms" as $1$ on the LHS), the general term is $$\frac{(-1)^k}{2^k k!}\frac{n!}{(n-2k)!}\frac{2^n n!}{(2n)!}\frac{(2n-2k)!}{2^{n-k}(n-k)!}=\frac{n!^2}{(2n)!}(-1)^k\binom{n}{k}\binom{2n-2k}{n}$$ (again, the "$1+{}$" part of the summation may be viewed as including the term with $k=0$).
And the key to a solution is indeed related to the Legendre polynomials, defined here by $$P_n(x)=\frac{1}{2^n n!}\frac{d^n}{dx^n}(x^2-1)^n$$ (Rodrigues' formula). If we expand $(x^2-1)^n$ and take the derivative termwise, we get $$P_n(x)=\frac{1}{2^n}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{k}\binom{2n-2k}{n}x^{n-2k},$$ so that the sum we're computing is equal to $\frac{2^n n!^2}{(2n)!}P_n(1)$.
Finally, $P_n(1)=1$, which is shown using $(x^2-1)^n=(x-1)^n(x+1)^n$ and Leibniz rule.