Please help prove
$$
\begin{align}
\frac{\pi}{2}&=\left(\frac{1}{2}\right)^{2/1}\left(\frac{2^{2}}{1^{1}}\right)^{4/(1\cdot 3)}\left(\frac{1}{4}\right)^{2/3}\left(\frac{2^{2}\cdot4^{4}}{1^{1}\cdot3^{3}}\right)^{4/(3\cdot 5)}\left(\frac{1}{6}\right)^{2/5}\left(\frac{2^{2}\cdot4^{4}\cdot6^{6}}{1^{1}\cdot3^{3}\cdot5^{5}}\right)^{4/(5\cdot 7)}\cdots\\[5pt]
&=\prod_{n=1}^{\infty}\left(\frac{1}{2n}\right)^{\frac{2}{2n-1}}\left(\prod_{k=1}^{n}\frac{\left(2k\right)^{2k}}{\left(2k-1\right)^{2k-1}}\right)^{\frac{4}{\left(2n-1\right)\left(2n+1\right)}}
\end{align}
$$
Here's my progress
$$
\begin{align}
\frac{\pi}{2}&=\prod_{n=1}^{\infty}\left(\frac{1}{2n}\right)^{\frac{2}{2n-1}}\left(\prod_{k=1}^{n}\frac{\left(2k\right)^{2k}}{\left(2k-1\right)^{2k-1}}\right)^{\frac{4}{\left(2n-1\right)\left(2n+1\right)}}\\[5pt]
&=\prod_{n=1}^{\infty}\left(\frac{1}{2n}\right)^{\frac{2}{2n-1}}\left(\prod_{k=1}^{n}\frac{\left(2k\right)^{2k}}{\left(2k-1\right)^{2k-1}}\right)^{\frac{2}{2n-1}-\frac{2}{2n+1}}\\[5pt]
&=\lim\limits_{m\to\infty}\frac{\prod_{n=1}^{m}\left(\frac{1}{2n}\right)^{\frac{2}{2n-1}}\left(\prod_{k=1}^{n}\frac{\left(2k\right)^{2k}}{\left(2k-1\right)^{2k-1}}\right)^{\frac{2}{2n-1}}}{\left(\prod_{k=1}^{m}\frac{\left(2k\right)^{2k}}{\left(2k-1\right)^{2k-1}}\right)^{\frac{2}{2m+1}}\prod_{n=2}^{m}\left(\prod_{k=1}^{n-1}\frac{\left(2k\right)^{2k}}{\left(2k-1\right)^{2k-1}}\right)^{\frac{2}{2n-1}}}\\[5pt]
&=\lim\limits_{m\to\infty}\frac{\prod_{n=1}^{m}\left(\frac{1}{2n}\right)^{\frac{2}{2n-1}}\left(\frac{\left(2n\right)^{2n}}{\left(2n-1\right)^{2n-1}}\right)^{\frac{2}{2n-1}}}{\left(\prod_{k=1}^{m}\frac{\left(2k\right)^{2k}}{\left(2k-1\right)^{2k-1}}\right)^{\frac{2}{2m+1}}}\\[5pt]
&=\lim\limits_{m\to\infty}\frac{\left(\prod_{n=1}^{m}\frac{2n}{2n-1}\right)^{2}}{\left(\prod_{k=1}^{m}\frac{\left(2k\right)^{2k}}{\left(2k-1\right)^{2k-1}}\right)^{\frac{2}{2m+1}}}\\[5pt]
&=\lim\limits_{m\to\infty}\frac{\pi\left(\frac{\Gamma(m+1)}{\Gamma(m+\frac12)}\right)^{2}}{\left(\prod_{k=1}^{m}\frac{\left(2k\right)^{2k}}{\left(2k-1\right)^{2k-1}}\right)^{\frac{1}{m}}}\\[5pt]
\iff \frac12&=\lim\limits_{m\to\infty}\frac{\left(\frac{\Gamma(m+1)}{\Gamma(m+\frac12)}\right)^{2}}{\left(\prod_{k=1}^{m}\frac{\left(2k\right)^{2k}}{\left(2k-1\right)^{2k-1}}\right)^{\frac{1}{m}}}
\end{align}
$$
I'm stuck here for now.
Edit:
Drawing from this post we apparently just need to show that
$$\left(\prod_{k=1}^{m}\frac{\left(2k\right)^{2k}}{\left(2k-1\right)^{2k-1}}\right)^{\frac{1}{m}}\sim 2m$$
Best Answer
I believe we can reduce this to Wallis's product. I haven't written out the full justification, but it ought to be straightforward.
Assume that we can rearrange the product arbitrarily. This is the step I haven't justified, but it ought t be easy, since everything in sight is positive. In particular, we could take the base $2$ logarithm of the partial products to get the partial summand of a sequence of positive terms.
Now if we pull out the expressions with a factor of $$\frac{(2n)^{2n}}{(2n-1)^{2n-1}},$$ we get $$\left(\frac1{2n}\right)^{2/(2n-1)}\prod_{k=n}^\infty\left(\frac{(2n)^{2n}}{(2n-1)^{2n-1}}\right)^{4/((2k-1)(2k+1))}\tag1$$ where I've also associated each of the terms of the form $\left(\frac1{2n}\right)^{2/(2n-1)}$ with one such expression.
The logarithm of the product in $(1)$ is $$ \log\left(\frac{(2n)^{2n}}{(2n-1)^{2n-1}}\right)\sum_{k=n}^\infty\frac4{(2n-1)(2n+1)}= \frac2{2n-1}\log\left(\frac{(2n)^{2n}}{(2n-1)^{2n-1}}\right) $$ because the sum telescopes.
Now $(1)$ becomes $$\left(\frac1{2n}\right)^{2/(2n-1)}\left(\frac{(2n)^{2n}}{(2n-1)^{2n-1}}\right)^{2/(2n-1)}= \left(\frac{2n}{2n-1}\right)^2,$$ and taking the product over all $n$ gives $$\prod_{n=1}^\infty\left(\frac{2n}{2n-1}\right)^2,$$ which is clearly equivalent to Wallis's product for $\frac\pi2$.