 Solution

Iterate the Gamma function recurrence to obtain, near $z=-m$ for $m\ge 0$,The numerator is holomorphic and equals $\Gamma (1)=1$ at $z=-m$. Exactly one denominator factor, $z+m$, vanishes; the product of the remaining factors there is $(-m)(-m+1)\cdots (-1)=(-1{)}^{m}m!$. Thus there is a simple pole, with Gamma function residue at a nonpositive integerFor $m=0$ the empty product is one. The same recurrence with a sufficiently large shift defines the continuation at every point away from these integers, so no further poles occur. At the poles, the recurrence is an identity of meromorphic functions, rather than an equality of finite values.