= Gamma function recurrence
{c}
{title2=$\Gamma(z+1)=z\Gamma(z)$}
The <Gamma function> obeys the <recurrence relation>
$$
\Gamma(z+1)=z\Gamma(z).
$$
Initially, for $\operatorname{Re}z>0$, <integration by parts> in the <Gamma integral> gives the short proof
$$
\Gamma(z+1)=\int_0^\infty t^ze^{-t}\,dt=\left[-t^ze^{-t}\right]_0^\infty+z\int_0^\infty t^{z-1}e^{-t}\,dt=z\Gamma(z).
$$
The boundary term vanishes at zero because $|t^z|=t^{\operatorname{Re}z}\to0$, 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> $n$.https://dlmf.nist.gov/5.5#E1[NIST DLMF, equation 5.5.1: Gamma recurrence]{ref}
Read backwards, the <Gamma function recurrence> defines
$$
\Gamma(z)=\frac{\Gamma(z+m)}{z(z+1)\cdots(z+m-1)},\qquad \operatorname{Re}(z+m)>0,
$$
where $m$ 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>.
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