How to prove that Legendre polynomials can be obtain by Gram-Schmidt process? If we start with the polynomials $1, x, ..., x^n$ on $[-1,1]$ and apply the Gram-Schmidt process with $\lt f\cdot g\gt =\int\limits_{-1}^{1} f(x)g(x)dx$. How can I prove that the new polynomials ${L_0,...,L_n}$ are Legendre polynomials: 
$$L_n(x)=\dfrac{1}{2^nn!}\dfrac{d^n}{dx^n}[(x^2-1)^n]$$ 
with the Gram-Scmidt process?
 A: Gram-Schmidt for the case of the Legendre polynomials can be naively performed by defining $p_n$ to be the $L^2$ coprojection of $x^n$ onto the span of $1,x,\dots,x^{n-1}$. 
Computing this coprojection using Gram-Schmidt gives us the recursion:
$$p_n(x)=x^n-\sum_{k=0}^{n-1} \frac{\int_{-1}^1 p_k(y) y^n dy}{\int_{-1}^1 p_k(y)^2 dy} p_k (x).$$
This works. But we can do better by choosing to project $x p_{n-1}(x)$ instead of $x^n$. Now $x p_{n-1}(x)$ still has degree $n$ and the leading coefficients are the same, so this will give us the same as projecting $x^n$ would. Thus we write:
$$p_n(x)=x p_{n-1}(x)-\sum_{k=0}^{n-1} \frac{\int_{-1}^1 y p_k(y) p_{n-1}(y) dy}{\int_{-1}^1 p_k(y)^2 dy} p_k(x).$$
Why is this any better? Because actually only two terms of the sum are nonzero! This is because $p_k$ is orthogonal to $p_j$ for all $j<k$, and is therefore orthogonal to all polynomials of degree $d<k$ as well. Therefore the inner product $(yp_k,p_{n-1})$ is zero as soon as $k+1<n-1$. (It is precisely the trick $(p_k,xp_{n-1})=(xp_k,p_{n-1})$ that is specific to the world of orthogonal polynomials. Essentially everything else we did here would work in an abstract inner product space.) Thus we get the "three term recurrence relation":
$$p_n(x)=x p_{n-1}(x)-\frac{\int_{-1}^1 y p_{n-1}(y)^2 dy}{\int_{-1}^1 p_{n-1}(y)^2 dy} p_{n-1}(x) - \frac{\int_{-1}^1 y p_{n-2}(y) p_{n-1}(y) dy}{\int_{-1}^1 p_{n-2}(y)^2} p_{n-2}(x).$$
By proving formulae for these integrals by induction, you can derive a general form for this three term recurrence relation. Standardizing by taking $P_n(x)=\frac{p_n(x)}{p_n(1)}$ gives the standard Legendre polynomials. (Note that the Legendre polynomials are not orthonormal in the sense of $L^2$.)
Now given the resulting recurrence for the Legendre polynomials, which turns out to be 
$$(n+1) P_{n+1}(x)=(2n+1)xP_n(x)-nP_{n-1}(x)$$
you can prove this recurrence given another expression for the Legendre polynomials, e.g. a Rodriguez formula like the one you posted. It then follows that the two are the same once you check that they satisfy the same initial conditions i.e. $P_0=1,P_1=x$.
