Evaluate $\prod_{n=1}^{\infty}\left(\frac{\Gamma\left(2^n+\frac{1}{2}\right)}{a^n\Gamma(2^n)}\right)^{2^{-n}}$ An interesting infinite product with the closed form,
How can we show show that the formula is correct
$$\prod_{n=1}^{\infty}\left(\frac{\Gamma\left(2^n+\frac{1}{2}\right)}{a^n\Gamma(2^n)}\right)^{2^{-n}}=\frac{8}{a^2}\cdot \frac{\sqrt{\pi}}{e^2}$$ 
Let 
$$\prod_{n=1}^{\infty}\left(\frac{\Gamma\left(2^n+\frac{1}{2}\right)}{a^n\Gamma(2^n)}\right)^{2^{-n}}=X$$
take the log
$$\sum_{n=1}^{\infty}\frac{1}{2^n}\log\frac{\Gamma\left(2^n+\frac{1}{2}\right)}{a^n\Gamma(2^n)}=\log X$$ 
$$\sum_{n=1}^{\infty}\frac{1}{2^n}\log\Gamma\left(2^n+\frac{1}{2}\right)-\sum_{n=1}^{\infty}\frac{1}{2^n}\log\Gamma\left(2^n\right)-\sum_{n=1}^{\infty}\frac{n}{2^n}\log a=\log X$$ 
$$\sum_{n=1}^{\infty}\frac{1}{2^n}\log\Gamma\left(2^n+\frac{1}{2}\right)-\sum_{n=1}^{\infty}\frac{1}{2^n}\log\Gamma\left(2^n\right)-\log a^2=\log X$$ 
$$\sum_{n=1}^{\infty}\frac{1}{2^n}\log\Gamma\left(2^n+\frac{1}{2}\right)-\sum_{n=1}^{\infty}\frac{1}{2^n}\log\Gamma\left(2^n\right)=\log(a^2X)$$ 
$$\sum_{n=1}^{\infty}\frac{1}{2^n}\log\Gamma\left(2^n+\frac{1}{2}\right)-\sum_{n=1}^{\infty}\frac{\log(2^n-1)!}{2^n}=\log(a^2X)$$ 
...

As I said in the comments, evaluate the finite product and then take the limit as the number of terms, $N \to \infty$.
First let's state up front that
$$\frac{\Gamma{\left ( n+\frac12 \right )}}{\Gamma{(n)}} = \sqrt{\pi} n \frac{(2 n)!}{2^{2 n} (n!)^2} $$
so that we can write the product out to $N$ terms as follows:
$$\left (\sqrt{\pi}
\left ( \frac{2}{a^2} \right)^1 \frac{\left (2^{2} \right )!}{2^{2^{2}} \left ( 2^1 \right )!^2} \right )^{1/2^1} \left (\sqrt{\pi}
\left ( \frac{2}{a^2} \right)^2 \frac{\left (2^{3} \right )!}{2^{2^{3}} \left ( 2^2 \right )!^2} \right )^{1/2^2} \cdots \left (\sqrt{\pi}
\left ( \frac{2}{a^2} \right)^N \frac{\left (2^{N+1} \right )!}{2^{2^{N+1}} \left ( 2^N \right )!^2} \right )^{1/2^N}$$
which can be simplified to a single term as follows.  First, the power of the $\sqrt{\pi}$ term is a geometric series: 
$$\frac12 + \frac1{2^2} + \cdots +\frac1{2^N} = 1-\frac1{2^N}$$ 
Then the power of the $2/a^2$ term is another well-known series:
$$\frac12 + \frac{2}{2^2} + \cdots +\frac{N}{2^N} = 2-\frac{N+2}{2^N}$$
Finally, we get cancellations of the $\left ( 2^{n+1} \right )!$ terms in the denominator, leaving one of the corresponding $\left ( 2^{n} \right )!$ in the denominator.  The product is then, exactly,
$$\left ( \sqrt{\pi} \right )^{1-2^{-N}} \left (\frac{2}{a^2} \right )^{2-(N+2) 2^{-N}} 2^{-(2 N+1)} \left ( \left ( 2^{N+1} \right )! \right )^{2^{-N}}$$
Now we consider the limit as $N \to \infty$.  Here we use Stirling and find that
$$\begin{align} \left ( \left ( 2^{N+1} \right )! \right )^{2^{-N}} &\approx (2 \pi)^{2^{-(N+1)}} 2^{2 N+2} 2^{(N+1) 2^{-(N+1)}} e^{-2}   \end{align} $$
so that, when plugged into the exact expression, and the limit as $N \to \infty$ is taken, produces as the product $8 \sqrt{\pi}/e^2 1/a^2$, as asserted.
