Quick way for proving $\Gamma(z) \neq 0$ for $z \in \mathbb C$

Question

Using the equation $\Gamma(z)\Gamma(1-z) = \frac \pi {\sin \pi z}$, this is easy to show that $\Gamma(z)$ never vanishes on $\mathbb C$.

Is there shorter way for showing $\Gamma$ is not zero ?

Answer 1

Going back to Gauss, we use the definition $$ \Gamma(x)=\lim_{n\to\infty}\frac{n!n^x}{(x+1)(x+2)\cdots (x+n))}$$ for $x\in \Bbb C\setminus\{0,-1,-2,\ldots\}$. From this we obtain $$\Gamma(z)=\frac1z\prod_{n=1}^\infty\left(1+\frac xn\right)^{-1}e^{x\ln\frac{n+1}n}. $$ The very fact that this converges in the sense of convergent infinite products tells us that $\Gamma(z)\ne 0$ except possibly at non-positive integers. The rest follows from $\Gamma(z+1)=z\Gamma(z)$.