Let $w$ be a positive continuous function and let $n$ be a nonnegative integer. Equip $\mathcal{P_n}(\mathbb{R})$ with the inner product $$ \langle p, q \rangle = \int_{0}^{1}p(x)q(x)w(x)dx.$$ Let $p_0, p_1, ..., p_n$ be an orthonormal basis for $\mathcal{P}_n(\mathbb{R})$ where each $deg(p_k) = k$. Show that $\langle p_k, p_k' \rangle = 0$ for each $k$, where $p_k'$ is the derivative.
I don't know where to begin with this. I was thinking of proving it arithmetically using the general formula of $p_k$ and $p_k'$ from Gram-Schmidt, but I was hoping that there is a more elegant solution.
EDIT: Added bottom explanation.
Hint: It suffices to observe that $\deg (p'_k) = k-1$. Thus $p'_k$ is a linear combination of $p_0,\ldots,p_{k-1}$.
Note that the special definition of the inner product is completely irrelevant. If $\langle -, - \rangle$ is any inner product on $\mathcal{P_n}(\mathbb{R})$, you can apply the Gram–Schmidt process to the basis $\{1, x, x^2,\ldots, x^n\}$ and obtain an orthonormal basis $\{p_0, p_1, p_2, \ldots , p_n\}$ such that $\operatorname{span} (p_0,\dots,p_i) = \operatorname{span} (1,\dots,x^i)$ for $i = 0,\ldots, n$. This implies that $\deg(p^i) = i$.