Gamma function recurrence ($\Gamma (z+1)=z\Gamma (z)$)

The Gamma function obeys the recurrence relationInitially, for $\mathrm{Re}z>0$, integration by parts in the Gamma integral gives the short proofThe boundary term vanishes at zero because $|t^z|=t^{\mathrm{Re}z}\to 0$, and at infinity because of exponential decay. Together with $\Gamma (1)={\int }_0^{\infty }{e}^{-t}dt=1$, the Gamma function recurrence immediately gives $\Gamma (n+1)=n!$ for every nonnegative integer $n$.^[ref]

Read backwards, the Gamma function recurrence defineswhere $m$ is a positive integer. These overlapping formulas agree by the Gamma function recurrence, giving the meromorphic continuation. Since $\Gamma (1)=1$, they give a simple pole at zero and then at every negative integer; Gamma function residue at a nonpositive integer records the residues. The recurrence relation therefore holds throughout the continuation as an identity of meromorphic functions.