# Gamma(z+1)=zGamma(z)

The following formula holds: $$\Gamma(z+1)=z\Gamma(z),$$ where $\Gamma$ denotes gamma.
Use integration by parts to compute $$\begin{array}{ll} \Gamma(z+1) &= \displaystyle\int_0^{\infty} \xi^z e^{-\xi} \mathrm{d}\xi \\ &= -\xi^z e^{-\xi}\Bigg|_0^{\infty}- \displaystyle\int_0^{\infty} z \xi^{z-1} e^{-\xi} \mathrm{d}\xi \\ &= z\Gamma(z), \end{array}$$ as was to be shown. █