How to show that $P_n(x)$ have n distinct roots

Question

Let $P_n : \mathbb R \to \mathbb R $ , $n\in \mathbb N$ be defined by
$P_n(x)=\frac{\displaystyle1}{\displaystyle2^n n!}\frac{\displaystyle d^n}{\displaystyle dx^n}[(x^2-1)^n]$
I need to show that $P_n(x)$ has exactly $n$ distinct roots in $(-1,1)$ I think it is important to use induction and Rolle's theorem.
To use induction, $P_1(x)=x$ and it is trivial that the root is $x=0$ in $(-1,1)$ , but the inductivity is problem.
I assumed that $P_n(x)$ has $n$ distinct roots, and yet I can't advance further.

Answer 1

Try to make a statement about $\frac{\displaystyle d^k}{\displaystyle dx^k}[(x^2-1)^n]$, i.e. show by induction on $k$ that $x=1$ and $x=-1$ are roots of multiplicity $n-k$ and tehre are $k$ simple roots in $(-1,1)$.