Proof. We use Leibniz’s formula for derivative of a product, to get \begin {align*} \frac {d^n}{dx^n}\left [(1 - x)^{\alpha + n}\, (1 + x)^{\beta + n}\right ] = & \sum ^n_{k=0} \frac {n!}{k!\, (n-k)!}\left \{\frac {d^k}{dx^k}(1 + x)^{\beta + n}\right \}\left \{\frac {d^{n-k}}{dx^{n-k}}(1 - x)^{\alpha + n}\right \}\\ = & \sum ^n_{k=0}\frac {n!}{k!\, (n-k)!}\, (\beta + n)(\beta + n -1)\, \cdots \, (\beta + n - k + 1)(1+x)^{\beta + n - k}\\ & \times (-1)^{n-k}\, (\alpha + n)(\alpha + n - 1)\, \cdots \, (\alpha + n - n + k + 1)(1 - x)^{\alpha + k}\\ = & \sum ^n_{k = 0} \frac {n!\, (-1)^{n-k}}{k!\, (n-k)!}\frac {\Gamma (\beta + n + 1)\, \Gamma (\alpha + n + 1)}{\Gamma (\beta + n - k + 1)\, \Gamma (\alpha + k + 1)}\, (1 + x)^{\beta + n - k}\, (1 - x)^{\alpha + k} \end {align*}

where we have used \(\Gamma (x + 1) = x\Gamma (x)\). Replacing in Rogrigues formula get \begin {align*} P_n^{(\alpha ,\beta )}(x) & = \frac {(-1)^n}{2^n\, n!}\, (1 - x)^{-\alpha }\, (1 + x)^{-\beta }\, \frac {d^n}{dx^n}\left [(1 + x)^{\beta + n}(1 - x)^{\alpha + n}\right ]\\ & = \sum ^n_{k=0}\frac {\Gamma (\alpha + n + 1)\, \Gamma (\beta + n + 1)\, \left (\frac {x - 1}{2}\right )^k\, \left (\frac {x + 1}{2}\right )^{n-k}}{\Gamma (\alpha + k + 1)\, \Gamma (\beta + n-k +1)\, k!\, (n-k)!}. \end {align*} □