TITLE: How does one prove Rodrigues' formula for Legendre Polynomials?
QUESTION [10 upvotes]: I am trying to prove that $\frac{1}{n!\space2^n}\frac{d^n}{dx^n}\{(x^2-1)^n\}=P_n(x)$, where $P_n(x)$ is the Legendre Polynomial of order n.
I've been told that the proof uses complex analysis, of which I know nothing, isn't there a proof with elementary methods? (If there isn't, I'm still interested in the other one).

REPLY [2 votes]: (The general formula of Legendre Polynomials is given by following equation:
$$
P_k(x)=\sum_{m=0}^{\frac{k}{2}|\frac{k-1}{2}}{\frac{(-1)^m(2k-2m)!}{2^km!(k-m)!}}\frac{1}{(k-2m)!} x^{k-2m}
$$
The Rodrigues' formula is:
$$ \frac{1}{2^kk!}\frac{d^k}{dx^k}[(x^2-1)^k] $$
The Binomial theorem is as follow:
$$(x+y)^k=\sum_{i=0}^{k}\frac{k!}{i!(k-i)!}x^{k-i}y^{i}$$ 
Then $$(x^2-1)^k=\sum_{i=0}^{k}\frac{k!}{i!(n-i)!}{(x^{2})}^{k-i}(-1)^{i}$$
$$\frac{1}{2^kk!}\frac{d^k}{dx^k}[(x^2-1)^k]=\frac{1}{2^kk!}\frac{d^k}{dx^k}\big[\sum_{i=0}^{k}\frac{k!}{i!(n-i)!}{(x^{2})}^{k-i}(-1)^{i}\big]$$
So
$$ =\frac{1}{2^kk!}\sum_{i=0}^{k}\frac{k!}{i!(n-i)!}\frac{d^k}{dx^k}(x)^{2k-2i}(-1)^{i} .....(1)$$ 
$$\frac{d^k}{dx^k}(x)^{2k-2i}=\frac{(2k-2i)!}{k-2i}x^{k-2i}......(2)$$
By compensate (2) into (1), we get:
$$
P_k(x)=\sum_{m=0}^{\frac{k}{2}|\frac{k-i}{2}}{\frac{(-1)^i(2k-2i)!}{2^ki!(k-i)!}}\frac{1}{(k-2i)!} x^{k-2i}
$$
Change dummy variable (i) to (m), we get the same general formula of Legendre Polynomials :
$$
P_k(x)=\sum_{m=0}^{\frac{k}{2}|\frac{k-1}{2}}{\frac{(-1)^m(2k-2m)!}{2^km!(k-m)!}}\frac{1}{(k-2m)!} x^{k-2m}
$$