TITLE: Proof the Legendre polynomial $P_n$ has $n$ distinct real zeros
QUESTION [9 upvotes]: 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 $(-1, 1)$.

REPLY [2 votes]: 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) $.