Prove that the Legendre polynomials $P_n$ have $n$ zeros in $(-1,1)$
I already proved that the Legendre polynomials $P_n$ have degree $n$ and any zero belong to $(-1,1)$, but Im stuck trying to prove that these polynomials have $n$ zeros.
I tried different induction hypothesis to build a proof but no one works.
The definition of the Legendre polynomials is
$$P_n(x):=\frac1{2^nn!}\partial^n[(x^2-1)^n],\quad n\in\Bbb N_{\ge 0}$$
I derived the alternative forms
$$\begin{align}P_n(x)&=2^{-n}\sum_{k=0}^n\binom{n}{k}^2(x+1)^{n-k}(x-1)^k\\&=\frac1{2^nn!}\sum_{k=\lceil n/2\rceil}^n\binom{n}{k}(2k)^{\underline n}(-1)^{n-k}x^{2k-n}\end{align}$$
where $(2k)^{\underline n}=\prod_{j=0}^{n-1}(2k-j)$ is a falling factorial. I have some useful identities to play as
$$\partial^n (fg)=\sum_{k=0}^n(\partial^k f)\cdot(\partial^{n-k}g)$$
$$\partial^{n+1}(x^2-1)^{n+1}=\partial^n[\partial (x^2-1)^n]=2n\partial^nx(x^2-1)^{n-1}$$
$$\partial^nx(x^2-1)^n=x\partial^n(x^2-1)^n+n\partial^{n-1}(x^2-1)^n$$
but as I said I cant get something useful from here. Some help will be very welcome, thank you.
P.D.: for this proof I cant use any integration theorem or so, this is a proof on the context of differentiability.
