 Solution

For $\mathrm{Re}z>0$, absolute convergence permits the two-dimensional gamma integralPut $u=rs$, $v=r(1-s)$, with $r>0$, $0<s<1$ and Jacobian determinant $r$. Separation of the integrals gives the beta-gamma identityThe beta integral is finite in this half-plane. Thus a zero of $\Gamma (z)$ would imply $\Gamma (z/2)=0$, and iteration would imply $\Gamma (z/2^m)=0$ for every positive integer $m$. But integration by parts gives $w\Gamma (w)=\Gamma (1+w)$, and continuity near one makes $\Gamma (1+w)\to 1$ as $w\to 0$. It cannot vanish for sufficiently small $w$, a contradiction. Therefore the Gamma function has no zeros in the right half-plane.

For any $z$ outside the nonpositive integers, choose an integer $n$ with $\mathrm{Re}(z+n)>0$. The Gamma function recurrence givesAt the excluded integers gamma has poles, with nonzero residues by the same recurrence. Hence the meromorphic gamma function has no zeros anywhere; at its poles it has no finite value to which a nonzero-value assertion could apply.