Chapter 11.5, Problem 6P

Prove that $\frac{d}{dp}\Gamma (p)={\displaystyle {\int }_0^{\infty }{x}^{p-1}}{e}^{-x}\ln xdx,$ $\frac{d^n}{d{p}^n}\Gamma (p)={\displaystyle {\int }_0^{\infty }{x}^{p-1}}{e}^{-x}{(\ln x)}^{n}dx.$