= Solution
For $\operatorname{Re}z>0$, absolute convergence permits the two-dimensional gamma integral
$$
\Gamma(z/2)^2=\int_0^\infty\int_0^\infty
e^{-(u+v)}u^{z/2-1}v^{z/2-1}\,du\,dv.
$$
Put $u=rs$, $v=r(1-s)$, with $r>0$, $0<s<1$ and <Jacobian determinant> $r$. Separation of the integrals gives the <beta--gamma identity>
$$
\boxed{\Gamma(z/2)^2=\Gamma(z)B(z/2,z/2)}.
$$
The 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 $\Gamma(1+w)\to1$ as $w\to0$. 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 $\operatorname{Re}(z+n)>0$. The <Gamma function recurrence> gives
$$
\Gamma(z)=\frac{\Gamma(z+n)}{z(z+1)\cdots(z+n-1)}\ne0.
$$
At the excluded integers gamma has poles, with nonzero residues by the same recurrence. Hence \b[the meromorphic <gamma function> has no zeros anywhere]; at its poles it has no finite value to which a nonzero-value assertion could apply.
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