 Solution

Let $x_1,\dots ,x_k$ be the distinct zeros in $(-1,1)$ at which $P_n$ changes sign, and put $Q(x)={\prod }_{j=1}^k(x-x_j)$. Then $P_{n}Q$ has constant sign and is nonzero except at finitely many points, soIf $k<n$, this contradicts the orthogonality of $P_n$ to the lower-degree polynomial $Q$. Hence $k\ge n$; degree $n$ forces exactly $n$ distinct, simple roots, all in $(-1,1)$.