Legendre polynomials have all their roots in (-1, 1)

Claim. For $0\le k\le n$, $Q_k=(x^2-1)^{n-k}R_k$, where $R_k$ is a polynomial of degree $k$ with $k$ distinct roots in $(-1,1)$.

Degree check. $Q_k$ has degree $2n-k$, so $\deg R_k=2n-k-2(n-k)=k$. Hence the $k$ roots are all the roots of $R_k$, and they are simple.

Start. $k=0$: $Q_0=(x^2-1)^n$ and $R_0=1$ has $0$ roots.

Inductive step. Suppose the claim holds for some $k<n$. Then $Q_k=(x-1)^{n-k}(x+1)^{n-k}R_k$ vanishes at the $k+2$ distinct points
$$-1<r_1<\dots<r_k<1,$$
because $n-k\ge1$. By Rolle's theorem applied to $Q_k$ on each of the $k+1$ intervals between consecutive points, $Q_{k+1}=Q_k'$ vanishes at some point in each of them: that gives $k+1$ distinct roots in $(-1,1)$.

Moreover, differentiating a product $(x^2-1)^{m}R$ with $m=n-k\ge1$ leaves a factor $(x^2-1)^{m-1}$, so $Q_{k+1}=(x^2-1)^{n-k-1}R_{k+1}$ for a polynomial $R_{k+1}$. By the degree count, $R_{k+1}$ has degree $k+1$, so the $k+1$ roots found are all its roots.

Conclusion. For $k=n$, $Q_n=R_n$ has $n$ distinct roots in $(-1,1)$.

Remark. A degree $n$ polynomial cannot have more than $n$ roots, so these are all of them. This is the key property of the Legendre polynomials in numerical integration (Gauss quadrature).