On proving $\Gamma (z+1)=4^z \prod _{k=1}^{\infty } \frac{\Gamma \left(\frac{z}{2^k}+\frac{1}{2}\right)}{\sqrt{\pi }}$

Question

Gradshteyn & Ryzhik offers two interesting identities: $$\Gamma (z+1)=4^z \prod _{k=1}^{\infty } \frac{\Gamma \left(\frac{z}{2^k}+\frac{1}{2}\right)}{\sqrt{\pi }}$$ $$\Gamma (z)=2 (\frac{z}{e})^z \prod _{k=1}^{\infty } B\left(2^{k-1} z,\frac{1}{2}\right)^{\frac{1}{2^k}}$$ I think that they can be proved by using $3$ functional equations and Weierstrass/Euler product formula of Gamma function, but haven't found the right way. Any kind of help is appreciated.

Answer 1

Both are consequences of the duplication formula for $\Gamma$ (as noted in the comments). For the first one, write $$\frac{\Gamma(z+1/2)}{\sqrt\pi}\left[=2^{1-2z}\frac{\Gamma(2z)}{\Gamma(z)}\right]=4^{-z}\frac{\Gamma(1+2z)}{\Gamma(1+z)},$$ and now the product telescopes: $$\prod_{k=1}^\infty\frac{\Gamma\left(\frac{z}{2^k}+\frac12\right)}{\sqrt\pi}=\prod_{k=1}^{\infty}4^{-2^{-k}z}\frac{\Gamma(1+2^{1-k}z)}{\Gamma(1+2^{-k}z)}=4^{-z\sum\limits_{k=1}^{\infty}2^{-k}}\frac{\Gamma(1+z)}{\Gamma(1)}=4^{-z}\Gamma(1+z).$$ Similarly, for the second identity, write $\mathrm{B}(z,1/2)=2^{2z-1}\Gamma^2(z)/\Gamma(2z)$, so that $$\prod_{k=1}^{n}\big[\mathrm{B}(2^{k-1}z,1/2)\big]^{2^{-k}}=\prod_{k=1}^{n}2^{z-2^{-k}}\frac{\big[\Gamma(2^{k-1}z)\big]^{2^{-(k-1)}}}{\big[\Gamma(2^{k}z)\big]^{2^{-k}}}=2^{nz-1+2^{-n}}\frac{\Gamma(z)}{\big[\Gamma(2^{n}z)\big]^{2^{-n}}},$$ and it remains to note [replace $m$ with $2^n$ below], using Stirling's formula, that $$\lim_{m\to\infty}m^{-z}\big[\Gamma(mz)\big]^{1/m}=\lim_{m\to\infty}m^{-z}\big[(mz)^{mz-1/2}e^{-mz}\sqrt{2\pi}\big]^{1/m}=(z/e)^z.$$