Legendre Polynomial orthogonality It is given that $P_n (x) = \frac{d^n}{dx^n}(x^2-1)^n $
I have to show that these polynomials are a system of orthogonal polynomials.
I started like this :
$ \int_{-1}^{1} \frac{d^n}{dx^n}(x^2-1)^n * \frac{d^m}{dx^m}(x^2-1)^m $
with partial integration I came to 
$ \left [ \frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n* \frac{d^m}{dx^m}(x^2-1)^m \right ]_{-1}^1 - \int_{-1}^1 \frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n * \frac{d^{m+1}}{dx^{m+1}}(x^2-1)^m dx $
So the first part is $0$ because of $|x|=1$  and the integral equals $0$ because the $(m+1)$ Derivation of a polynomial of order $m$ is $0$ ?
I am not sure at all if this is right, can someone tell me if it's okay or if I cant do it this way?
 A: The following derivation is self-contained, in the sense that it does not leverage properties of Legendre polynomials.
To facilitate notations, let $Q_n(X) = (X^2-1)^n$
and let $Q_n^{(n)}$ denote the $n$-th derivative of $Q_n$, so that $P_n = Q_n^{(n)}$.
Let $1\leq n< m$ and note that
$$\langle P_n , P_m \rangle = \int_{-1}^1 Q_n^{(n)}(x) Q_m^{(m)}(x) dx = \Big[Q_n^{(n)}(x) Q_m^{(m-1)}(x)\Big]_{-1}^1 - \int_{-1}^1 Q_n^{(n+1)}(x) Q_m^{(m-1)}(x) dx.$$
Since $m\geq 1$, $1$ and $-1$ are roots of $Q_m$, each with multiplicity $m$ hence
$$Q_m^{(m-1)}(-1) = Q_m^{(m-1)}(1)=0.$$
If $n=1$ we stop here, otherwise we
integrate by parts repeatedly until $Q_m^{(m-n)}$ appears and we find
$$\langle P_n , P_m \rangle = (-1)^n \int_{-1}^1 Q_n^{(2n)}(x) Q_m^{(m-n)}(x) dx,$$
since for every $k\in \{2,\ldots,n\}$, $0\leq m-k\leq m-1 \implies Q_m^{(m-k)}(-1) = Q_m^{(m-k)}(1)=0$.
The degree of $Q_n$ is $2n$, thus $Q_n^{(2n)}$ is a constant polynomial: $Q_n^{(2n)} = (2n)!$, hence
$$\langle P_n , P_m \rangle = (-1)^n (2n)! \int_{-1}^1 Q_m^{(m-n)}(x) dx = 
(-1)^n (2n)! 
\Big[Q_m^{(m-n-1)}(x)\Big]_{-1}^1.
$$
Since $0\leq m-n-1\leq m-1$, this last bracket is $0$, and finally $\langle P_n , P_m \rangle = 0$.
The same technique yields the value of $\|P_n\|$.
