= Solution
Fix a <Hankel contour> with a positive-radius circle around zero and a branch $-\pi<\arg t<\pi$, traversing the lower bank inwards and the upper bank outwards. Write its numerator as $I(z)=\int_Ht^{z-1}e^t\,dt$. For fixed circle radius, $I$ is an <entire function> of $z$: the finite contour pieces avoid zero, and the tails converge uniformly on compact $z$ sets by exponential decay, also after differentiation with respect to $z$.
At $z=-n$, the two straight-bank contributions cancel because $t^{-n-1}e^t$ is single-valued. The remaining circle is positively oriented, so the <residue theorem> and the <exponential series> give
$$
I(-n)=2\pi i\operatorname{Res}_{t=0}(t^{-n-1}e^t)=\frac{2\pi i}{n!}.
$$
Since $\sin(\pi z)=(-1)^n\pi(z+n)+O((z+n)^3)$, the quotient has the <simple pole>
$$
\boxed{\operatorname{Res}_{z=-n}\Gamma(z)=\frac{(-1)^n}{n!},\qquad n\geq0.}
$$
For <cancellation of positive-integer gamma singularities on a Hankel contour>, at a positive <integer> $n$, the integrand $t^{n-1}e^t$ is entire and its bank contributions again cancel; its circular integral is zero. Hence $I(n)=0$. The numerator is analytic near $n$, while the denominator has a simple zero with nonzero derivative. Factoring $I(z)=(z-n)J(z)$ therefore cancels that zero, proving \b[there is no <pole> at a positive <integer>] using only this representation. Cancellation of the numerator, not merely the zero of the sine, decides the issue. One can also shrink the contour for $\Re z>0$, obtaining $I(z)=2i\sin(\pi z)\int_0^\infty u^{z-1}e^{-u}\,du$ and hence the finite value $(n-1)!$.
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