Gamma(z+1)=zGamma(z)
From specialfunctionswiki
Theorem: $$\Gamma(x+1)=x\Gamma(x), \quad x>0,$$ where $\Gamma$ denotes the gamma function.
Proof: Use integration by parts to compute $$\begin{array}{ll} \Gamma(x+1) &= \displaystyle\int_0^{\infty} \xi^x e^{-\xi} \mathrm{d}\xi \\ &= -\xi^x e^{-\xi}|_0^{\infty} \displaystyle\int_0^{\infty} x \xi^{x-1} e^{-\xi} \mathrm{d}\xi \\ &= x\Gamma(x), \end{array}$$ as was to be shown. █
