We have that $p(x) = (x^2-1)^m$ Note we need $m >2$.
Taking derivatives we have
$p'(x)=2xm(x^2-1)^{m-1}$
$p''(x)=4x^2m(m-1)(x^2-1)^{m-2}+2m(x^2-1)^{m-1}$
Rolle's Theorem asserts that if $f(x)$ is continuous in $I=[a,b]$ and $f(a)=f(b)$, then there is a $c \in I$ such that
$f'(c)=0$
We have that $f(-1) = f(1)$, so there is a $c$ such that $f'(c)=0$. We can immediatelly see such $c$ is $c=0$.
Moreover, we see that $f'(-1) = f'(1)=0$, but since $f'(0)=0$, we can assert that there exists $c_1 \in (-1,0)$ and $c_2 \in (0,1)$ such that $f''(c_1)=f''(c_2)=0$. Note that if $m=1,2$ the test would fail, since the first (third) derivative is a line.
You want to prove now that for $k=1,2,3,\dots ,m$, it is the case $p^{(k)}(c_i)=0$ for $i=1,2,\dots, m \in (-1,1) $. The induction hypothesis would then be:
$\mathcal H.$ It is true that $p^{(k)}(c_i)=0$ for $i=1,2,\dots,k \in (-1,1) $
This means that $p^{(k)}(c_1) = p^{(k)}(c_2) = \cdots = p^{(k)}(c_i)=0$. Since we also have two more roots, namely $1$ and $-1$, then we can assert by Rolle's Theorem, that $p^{(k+1)}(x)$ will have $n=k+2-1=k+1$ roots. Since $k=2$ is true, this completes the induction phase.