Claim
Let $P_n: \Bbb R \mapsto \Bbb R $, $n \in \Bbb N\cup\{0\}$
$P_n(x):={1\over{2^n{n!}}}{d^n\over dx^n}[(x^2-1)^n]$, then $P_n$ has n distinct roots in ]-1, 1[
Proof
$1$. Base
When $n=1$, $P_1(x) = {1 \over 2}{d \over dx}[x^2-1]=x$ and $P_1(x)$ has $1$ distinct root. Thus,
$Claim$ holds for the case of $n=1$.
$2$. Now assume the $Claim$ holds for the case of n=k. Then,
$P_k(x):={1\over{2^k{k!}}}{d^k\over dx^k}[(x^2-1)^k]$, then $P_k$ has k distinct roots in ]-1, 1[ .
Since $P_k(1)=P_k(-1)=0$, there $\exists c \in \Bbb N\cup\{0\}$ s.t. ${d \over dx}P_k(c)=0$ in $]-1,1[$ (by Rolle's Theorem).
Let $c$ be the root of $P_k(x)$. Then
$P_k(c)={1\over{2^k{k!}}}{d^k\over dc^k}[(c^2-1)^k]=0$
Question From abvoe proof, I had stucked in the middle of showing that $c$ is not one of the $n$ distinct roots of $P_k(x)$
any adivce or hint to proceed?
Question' Or, following the comment, I had found
$P_{k+1} = {1\over{2^k{k!}}}{d^k\over dx^k}[(x^2-1)^kx]$ which is slightly different from $P_k(x):={1\over{2^k{k!}}}{d^k\over dx^k}[(x^2-1)^k]$.
How could I use the fact that $P_k(x)$ has n distinct root and conclude $P_{k+1}(x)$ has one more root other than that of $P_k(x)$?