Showing orthogonality of Legendre polynomials using Rodrigues' formula 
Use Rodrigues’ formula to show that $I_l = \int_{-1}^1 P_l(x)P_l(x)dx = \frac{2}{2l +1}$.

Using Rodrigue's formula, we have 
$$I_l = \frac{1}{2^{2l}(l!)^2} \int_{-1}^1 \left[\frac{d^l(x^2-1)^l}{dx^l}\right]\left[\frac{d^l(x^2-1)^l}{dx^l}\right]dx$$
Now, it is in these below steps I'm having trouble with. Repeated integration by parts, with all boundary terms vanishing, reduces this to
$$I_l = \frac{(-1)^l}{2^{2l}(l!)^2} \int_{-1}^1 (x^2-1)^l \frac{d^{2l}(x^2-1)^l}{dx^{2l}}dx$$
or, $$I_l = \frac{(2l)!}{2^{2l}(l!)^2} \int_{-1}^1 (1 - x^2)^l dx$$
I could not find any literature that explains the above steps. 
Thank You.
P.S. - This was my first time writing in Mathjax.
 A: The proof works by mathematical induction, which is a basic mathematical technique.
Start with the integral you gave, and integrate by parts once:
\begin{align}
I_l 
& = 
\frac{1}{2^{2l}(l!)^2} \int_{-1}^1 \left[\frac{d^l(x^2-1)^l}{dx^l}\right]\left[\frac{d^l(x^2-1)^l}{dx^l}\right]dx
\\ & =
\frac{1}{2^{2l}(l!)^2} \left[
\left. \left[\frac{d^{l-1}(x^2-1)^l}{dx^{l-1}}\right]\left[\frac{d^{l}(x^2-1)^l}{dx^{l}}\right] \right|_{-1}^1
-
\int_{-1}^1 \left[\frac{d^{l-1}(x^2-1)^l}{dx^{l-1}}\right]\left[\frac{d^{l+1}(x^2-1)^l}{dx^{l+1}}\right]dx
\right]
\\ & =
\frac{(-1)}{2^{2l}(l!)^2}
\int_{-1}^1 \left[\frac{d^{l-1}(x^2-1)^l}{dx^{l-1}}\right]\left[\frac{d^{l+1}(x^2-1)^l}{dx^{l+1}}\right]dx
,
\end{align}
where the boundary terms vanish, because $(x^2-1)^l = (x+1)^l (x-1)^l$ has $l$-fold zeros at both endpoints, and the $(l-1)$-fold derivative leaves one zero on each end.
This suggests the intermediate result for the inductive step: the claim that
\begin{align}
I_l 
& = 
\frac{(-1)^k}{2^{2l}(l!)^2}
\int_{-1}^1 \left[\frac{d^{l-k}(x^2-1)^l}{dx^{l-k}}\right]\left[\frac{d^{l+k}(x^2-1)^l}{dx^{l+k}}\right]dx
,
\end{align}
for all $k=0,1,\ldots, l$. To prove this claim via induction, we assume that it's true for some $k$, and then we integrate by parts again:
\begin{align}
I_l 
& = 
\frac{(-1)^k}{2^{2l}(l!)^2}
\int_{-1}^1 \left[\frac{d^{l-k}(x^2-1)^l}{dx^{l-k}}\right]\left[\frac{d^{l+k}(x^2-1)^l}{dx^{l+k}}\right]dx
\\ & =
\frac{(-1)^{k}}{2^{2l}(l!)^2} \left[
\left. \left[\frac{d^{l-k-1}(x^2-1)^l}{dx^{l-k-1}}\right]\left[\frac{d^{l+k}(x^2-1)^l}{dx^{l+k}}\right] \right|_{-1}^1
-
\int_{-1}^1 \left[\frac{d^{l-k-1}(x^2-1)^l}{dx^{l-k-1}}\right]\left[\frac{d^{l+k+1}(x^2-1)^l}{dx^{l+k+1}}\right]dx
\right]
\\ & =
\frac{(-1)^{k+1}}{2^{2l}(l!)^2}
\int_{-1}^1 \left[\frac{d^{l-k-1}(x^2-1)^l}{dx^{l-k-1}}\right]\left[\frac{d^{l+k+1}(x^2-1)^l}{dx^{l+k+1}}\right]dx
,
\end{align}
where the boundary terms vanish for the same reason as above. This is exactly the same claim with $k$ replaced by $k+1$, which completes the proof by induction.
The desired result,
$$I_l = \frac{(-1)^l}{2^{2l}(l!)^2} \int_{-1}^1 (x^2-1)^l \frac{d^{2l}(x^2-1)^l}{dx^{2l}}dx,$$
then follows as the special case $k=l$. Finally, to get to
$$
I_l = \frac{(2l)!}{2^{2l}(l!)^2} \int_{-1}^1 (1 - x^2)^l dx
$$
you simply need to realize that $(x^2-1)^l$ is a polynomial of degree $2l$ with leading coefficient $1$, and that under a $2l$-fold derivative the only thing that will survive is $(2l)!$ times that leading coefficient.
