Gamma function identity involving 5/4, 3/4, 1/4 and pi

Question

How can I prove this identity: $$\frac{4\pi\sqrt{2}\Gamma\Big(\frac{5}{4}\Big)}{\Gamma\Big(\frac{3}{4}\Big)}-\Gamma\Big(\frac{1}{4}\Big)^2 = 0?$$ I found it while calculating $$\prod_{i=0}^\infty\frac{(4i+3)(4i+4)}{(4i+2)(4i+5)}$$ which WolframAlpha found was half of the lemniscate constant $$\frac{\Gamma(\frac{1}{4})^2}{2\sqrt{2\pi}}$$ as well as $$\frac{\sqrt{\pi}\cdot\Gamma(\frac{5}{4})}{\Gamma(\frac{1}{4})}.$$

Answer 1

Let $a:=\Gamma(1/4)$ so $\Gamma(3/4)=\tfrac{\pi\csc(\pi/4)}{a}=\tfrac{\pi\sqrt{2}}{a}$ and$$\frac{4\pi\sqrt{2}\Gamma(5/4)}{\Gamma(3/4)}=\frac{\pi\sqrt{2}a}{\pi\sqrt{2}/a}=a^2.$$