By means of the change of variable $u=r s, v=r(1-s)$ in a suitable double integral, or otherwise, show that for $\operatorname{Re} z>0$

$$\left[\Gamma\left(\frac{1}{2} z\right)\right]^{2}=B\left(\frac{1}{2} z, \frac{1}{2} z\right) \Gamma(z)$$

Deduce that, if $\Gamma(z)=0$ for some $z$ with $\operatorname{Re} z>0$, then $\Gamma\left(z / 2^{m}\right)=0$ for any positive integer $m$.

Prove that $\Gamma(z) \neq 0$ for any $z$.