= Solution
Iterate the <Gamma function recurrence> to obtain, near $z=-m$ for $m\geq0$,
$$
\Gamma(z)=\frac{\Gamma(z+m+1)}{z(z+1)\cdots(z+m)}.
$$
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 integer>
$$
\boxed{\operatorname*{Res}_{z=-m}\Gamma(z)=\lim_{z\to-m}(z+m)\Gamma(z)=\frac{(-1)^m}{m!}.}
$$
For $m=0$ 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.
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