Trying to prove that: $$\int_{-1}^1 P_n(x)P_k(x)\,dx=\frac{-1}{2^nn!} \int_{-1}^1 P_k'(x) \frac{d^{n-1}}{dx^{n+1}}(x^2-1)^n \,dx$$
I know that by the Rodriguez formula, $$P_n(x)=\frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2-1)^n$$
Would I just plug Rodriguez into the original integral? I tried that and have gotten to: $$\frac{1}{2^nn!}\int_{-1}^1 \frac{d^n}{dx^n}(x^2-1)^nP_k(x)\,dx$$
Any pointers on how to go about this?