 Parity of orthogonal polynomials for an even weight

For a positive even weight on a symmetric interval, any orthogonal polynomial of degree $k$ has the parity of $k$. Induct on $k$. The reflected polynomial $q_k(x)=(-1{)}^k{p}_k(-x)$ has the same leading coefficient. By substitution $x\mapsto -x$ in the weighted inner product and the inductive parity of all lower polynomials, it is orthogonal to them. Hence $q_k-p_k$ has lower degree and is orthogonal to the entire space containing it. Its squared norm is zero, forcing $q_k=p_k$. The base case is a constant. This proof needs neither monic nor unit-norm normalization; it only compares polynomials with the same leading coefficient.