Prove that $n$-th Jacobi polynomial $P_n^{(\alpha, \beta)}(x)$ with parameter $(\alpha, \beta)$ defined by
$$P_n^{(\alpha, \beta)}(x)=\frac{(-1)^n}{2^nn!}(1-x)^{-\alpha}(1+x)^{-\beta}\frac{d^n}{dx^n}\left[(1-x)^{\alpha}(1+x)^{\beta}(1-x^2)^n\right]$$
is equal to
$$\sum_{s=0}^n \binom{n+\alpha}{s}\binom{n+\beta}{n-s}\left(\frac{x-1}{2}\right)^{n-s}\left(\frac{x+1}{2}\right)^s.$$
I tried to arrive at the sum by differentiation using Leibniz's rule. But the calculation is super messy and I got nowhere. I also have tried to look up on the internet because I think this equality is standard but I cant find a proof.
You are in the right track. $$ (1-x)^{\alpha}(1+x)^{\beta}(1-x^2)^n=(1-x)^{\alpha+n}(1+x)^{\beta+n}, $$ and apply Leibniz's rule with $f(x):=(1-x)^{\alpha+n}$ and $g(x):=(1+x)^{\beta+n}$. After some calculation you will obtain the identity.