= Solution
Iterating the <Gamma function recurrence> gives a <meromorphic continuation>
$$
\Gamma(z)=\frac{\Gamma(z+m+1)}{z(z+1)\cdots(z+m)},\qquad m=0,1,2,\ldots.
$$
Near $z=-m$, 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 $(-1)^m m!$. Therefore
$$
\Gamma(z)\sim\frac{(-1)^m}{m!}\frac1{z+m}.
$$
Thus $\boxed{z=0,-1,-2,\ldots\text{ are all simple poles}}$. More precisely, the <Gamma function residue at a nonpositive integer> is $\boxed{\operatorname{Res}_{z=-m}\Gamma(z)=(-1)^m/m!}$. 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>.
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