Solution
= Solution
For any $n$ with $\operatorname{Re}z>-n$, define
$$
F(z)=\frac{F(z+n)}{z(z+1)\cdots(z+n-1)}.
$$
The definitions agree on overlaps by the recurrence, giving a meromorphic continuation to $\mathbb C$. Its only singularities are simple poles at $0,-1,-2,\ldots$. Since $F(1)=1$, repeated use of the recurrence gives
$$
\operatorname*{res}_{z=-n}F(z)=\frac{(-1)^n}{n!}.
$$
These residues are nonzero, so all the listed singularities really are simple poles.