 Solution

Iterating the Gamma function recurrence gives a meromorphic continuationNear $z=-m$, the numerator is analytic and tends to $\Gamma (1)=1$. Exactly one denominator factor, $z+m$, vanishes, while all other factors tend to the nonzero product $(-1{)}^{m}m!$. ThereforeThus $\boxed{z=0,-1,-2,\dots \text{ are all simple poles}}$. More precisely, the Gamma function residue at a nonpositive integer is $\boxed{{\mathrm{Res}}_{z=-m}\Gamma (z)=(-1{)}^m/m!}$. Taking $m$ large enough to put $z+m+1$ in the original right half-plane also shows that this continuation introduces no other finite poles.