Why does $\Gamma(1-z)\Gamma(z) = \frac{\pi}{\sin \pi z}$ imply $\Gamma(z) \not = 0$

Question

I'm reading on the extension of $\Gamma$ to the complex plane and there is written:

Corollary

$$\Gamma(z) \not = 0 \qquad \forall z \in \mathbb{C}\setminus\{0,-1,-2, \dots\}$$

Proof

$(\forall z \in \mathbb{C}\setminus\mathbb{Z}) \quad \Gamma(1-z)\Gamma(z) = \dfrac{\pi}{\sin \pi z}$ implies this. For $n\in \mathbb{N}$ is $\Gamma(n+1) = n! \not = 0$.

Why?

Can someone explain why that is true? Why does $\Gamma(1-z)\Gamma(z) = \dfrac{\pi}{\sin \pi z}$ imply $\Gamma(z) \not = 0$?

Answer 1

If $\Gamma(z_0)=0$ then

$$0 = \Gamma(1-z_0)\Gamma(z_0) = \dfrac{\pi}{\sin(\pi z_0)}$$

And $\dfrac{1}{z}\neq 0$ for any $z\in\mathbb{C}$

Answer 2

The RHS is nonzero, so no factor of the LHS can be zero.