Proof the Legendre polynomial $P_n$ has $n$ distinct real zeros I need a proof to show that the inequality $m < n$ leads to a contradiction and $P_n$ has $n$ distinct real roots, all of which lie in the open interval $(-1, 1)$.
 A: By Rodrigues formula for Legendre polynomials,
$$\displaystyle P_n(x) = \frac{1}{2^nn!}\frac{d^n}{dx^n} (x^2 - 1)^{n}\tag{*1}$$
$P_n(x)$ is the $n^{th}$ derivative of a polynomial with roots at $-1$ and $1$.
Repeat applying Gauss-Lucas theorem $n$ times, we know all roots of $P_n(x)$ lie on the closed line segment
$[-1,1]$ in the complex plane. By direct expansion of $(*1)$, one can check that
$P_n(\pm 1) \ne 0$, so the roots of $P_n(x)$ are all real and belongs to $(-1,1)$.
Since Legendre polynomials are solutions of the Legendre's differential equation:
$$\frac{d}{dx}\left[(1-x^2)\frac{d}{dx}P_n(x)\right] + n(n+1)P_n(x) = 0$$
which is a $2^{nd}$ order ODE, $P_n(x)$ cannot have any double roots. To see this,
let's say $P_n(x)$ has a double root at $\alpha \in (-1,1)$. 
$P_n(x)$ will then be a solution of following initial value problem:
$$\frac{d}{dx}\left[(1-x^2)\frac{d}{dx}y(x)\right] + n(n+1)y(x) = 0,\quad
\begin{cases}y(\alpha) = 0,\\y'(\alpha) = 0\end{cases}$$
Apply Picard-Lindelöf theorem to this $2^{nd}$ order ODE, the "uniqueness" part of the theorem tell us $P_n(x)$ vanishes over some neighbor of $\alpha$. Since $P_n(x)$ is not identically zero, this is impossible. As a consequence, all roots of $P_n(x)$ are simple.
A: Let Legendre polynomial be $$ f_n(x)=\frac{1}{2^n n!}\frac{d^{n}}{d x^n}(x^2-1)^n .$$
Note that $ (x^2-1)^n $ has roots $ -1, 1 $ each with multiplicity $ n $, and $$ (x^2-1)\Big|\frac{d^m}{d x^{m}}(x^2-1)^n, \, \text{when}\ m<n  .$$
Thus, $ -1, 1 $ are always roots of $ \frac{d^{m}}{d x^m}(x^2-1)^n $. And by Rolle's theorem, we know that there are $ n-1 $ distinct zeros $ x_1, x_2, \cdots, x_{n-1} $ in the interval $ (-1, 1) $ for the polynomial $ \frac{d^{n-1}}{d x^{n-1}}(x^2-1)^n $. Since $ -1, 1 $ are also zeros of $ \frac{d ^{n-1}}{d x^{n-1}}(x^2-1)^n $. Use Rolle's theorem once again, we get $ n $ distinct zeros in $ (-1, 1) $ and since $ \deg (\frac{d^{n}}{d x^{n}}(x^2-1)^n)=n $, we have found all the distinct roots of $ f_n(x) $ in $ (-1, 1) $.
A: From the wording of your question I guess you are looking for something like this:
The Legendre polynomial $P_n$ with $n>0$ has $n$ simple roots in $(-1, 1).$
Proof by contradiction: Assume  $P_n$ has $m$ with $0 \le m \lt n\;$ 
pairwise different zeroes  $x_1, x_2, \dots x_m$ of odd multiplicity in $(-1,1),\;$ i.e. $P_n$ changes sign in $x_i$, and consider the polynomial $Z_n(x) = (x-x_1)\cdot (x-x_2) \dots (x-x_m).\;$ Because $P_n$ is orthogonal to $P_0$ the intgeral $\int_{-1}^{1} P_n(x) dx$ vanishes and $P_n$ has at least one zero of odd multiplicity in $(-1,1),\,$ i.e.  $m>0.\;$  
The polynomial $Z_n(x) P_n(x)$ does not change sign in $(-1,1)$ and therefore $I_n = \int_{-1}^{1} Z_n(x) P_n(x) dx \ne 0.\;$ On the other hand $Z_n$ has degree $m$ and  can be written as a linear combination $Z_n = \sum_{k=0}^m c_k P_k(x)$. Thus we get the contradiction by means of the orthogonality of the $P_k$:
\begin{align}
0 \ne I_n &= \int_{-1}^{1} Z_n(x) P_n(x) dx\\
&= \int_{-1}^{1}  P_n(x) \sum_{k=0}^m c_k P_k(x) dx\\
&= \sum_{k=0}^m c_k \int_{-1}^{1}  P_n(x) P_k(x) dx\\
&= 0 
\end{align}
This shows that $P_n$ has $n$ pairwise different zeroes of odd multiplicity, and because $P_n$ has at most $n$ zeroes, the zeroes $x_1,\dots, x_n$ are simple.
Note that this kind of proof can be applied to general orthogonal polynomials over intervals $(a,b)$ with weighting function $w(x) \ge 0$.
