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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 $I(z)={\int }_H{t}^{z-1}{e}^{t}dt$. 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 giveSince $\sin (\pi z)=(-1{)}^n\pi (z+n)+O((z+n{)}^3)$, the quotient has the simple poleFor 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 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 $(n-1)!$.