I am trying to prove the Rodrigues formula for the Legendre polynomials from the power series recursion relation (obtained through the Frobenius method). On page 3 of this article, I can follow the author up to the following step
... The polynomial solution is therefore $$y(x)=\sum_{k=0}^{[n/2]}\frac{(-1)^k}{2^kk!}\frac{n(n-1)\cdots (n-2k+1)}{(2n-1)(2n-3)\cdots (2n-2k+1)}c_n x^{n-2k} \tag{1}$$ where $n\in \mathbb{Z}_+$ and $[]$ is the floor (greatest-integer) function.
But I can not for the life of me figure out how they arrive at the expression $P_n(x)$ (Legendre polynomials) with the simple substitution for $c_n$ that the author claims to make. More precisely, the author claims that for the following choice of $c_n$,
$$c_n=\frac{(2n)!}{2^n (n!)^2}$$
The polynomial (1) written above can be simplified to,
$$y(x)\rightarrow P_n(x)=\frac{1}{2^n}\sum_{k=0}^{[n/2]}\frac{(-1)^k\color{red}{(2n-2k)!}}{k!\color{red}{(n-k)!}(n-2k)!}x^{n-2k}$$
I have colored the terms that I don't understand the origin of in red. Could you please help me find how those terms get there?
Since we have $$n(n-1)\cdots (n-2k+1)=\frac{n!}{(n-2k)!}$$ and $$\begin{align}&(2n-1)(2n-3)\cdots (2n-2k+1)\\&=\frac{(2n-1)!}{(2n-2k)!}\cdot\frac{1}{(2n-2k+2)(2n-2k+4)\cdots (2n-2)}\\&=\frac{(2n-1)!}{(2n-2k)!}\cdot \frac{1}{2^{k-1}(n-k+1)(n-k+2)\cdots (n-1)}\\&=\frac{(2n-1)!}{(2n-2k)!}\cdot \frac{1}{2^{k-1}}\cdot\frac{(n-k)!}{(n-1)!}\end{align}$$ we have $$\begin{align}&\frac{(-1)^k}{2^kk!}\frac{n(n-1)\cdots (n-2k+1)}{(2n-1)(2n-3)\cdots (2n-2k+1)}c_n\\&=\frac{(-1)^k}{2^kk!}\cdot\frac{\color{red}{n(n-1)\cdots (n-2k+1)}}{\color{green}{(2n-1)(2n-3)\cdots (2n-2k+1)}}\cdot\frac{(2n)!}{2^n (n!)^2}\\&=\frac{(-1)^k}{2^kk!}\cdot \color{red}{\frac{n!}{(n-2k)!}}\cdot\color{green}{\frac{(2n-2k)!\cdot 2^{k-1}\cdot (n-1)!}{(2n-1)!(n-k)!}}\cdot\frac{(2n)!}{2^n (n!)^2}\\&=\frac{(-1)^k(2n-2k)!}{2^nk!(n-k)!(n-2k)!}\cdot\frac{(n-1)!(2n)!}{2\cdot(2n-1)!n!}\\&=\frac{(-1)^k(2n-2k)!}{2^nk!(n-k)!(n-2k)!}\end{align}$$