Verify this finite product $\prod_{r=1}^{2^k}\Gamma\left(1+\frac{r}{2^k}\right)^{(-1)^r}$

Question

How can we show that,

$$\prod_{r=1}^{2^k}\Gamma\left(1+\frac{r}{2^k}\right)^{(-1)^r}=\frac{(2^k-2)!!}{(2^k-1)!!}\cdot \frac{2^k}{\sqrt{2^{k-1}\pi}}$$

Where $\Gamma(n)$; Gamma function

Answer 1

$\displaystyle f(n):=\prod\limits_{r=1}^n\Gamma\left(1+\frac{r}{n}\right)\hspace{3cm}$ See GammaMultiplication.

$\displaystyle \prod\limits_{r=1}^{2^k}\Gamma\left(1+\frac{r}{2^k}\right)^{(-1)^r}=\frac{f(2^{k-1})^2}{f(2^k)}$

The rest is routine.