According to the Wikipedia article about Pi, if we set $z=\frac{1}{2}$ to $$\Gamma (z)=\dfrac{e^{-\gamma z}}{z}\displaystyle\prod_{n=1}^\infty \dfrac{e^{z/n}}{1+z/n}$$ and square the result, we get the Wallis product for $\pi$.
In my attempt, I ended up with $$\displaystyle\prod_{n=1}^\infty \dfrac{4n^2e^{1/n}}{(2n+1)^2}=\dfrac{\pi e^\gamma}{4},$$ but I don't know what to do next.
Note that $$\gamma=\lim_{n\to\infty}\left(1+\frac12+\cdots+\frac1n-\ln n\right)$$ and so $$e^{\gamma}=\lim_{n\to\infty}\frac{e^1e^{1/2}\cdots e^{1/n}}n.$$ Thus $$\prod_{n=1}^\infty \dfrac{4n^2e^{1/n}}{(2n+1)^2} =e^{\gamma}\lim_{N\to\infty}N\prod_{n=1}^N \frac{(2n)^2}{(2n+1)^2} =\frac{e^\gamma}2\lim_{n\to\infty}\frac21\frac23\frac43\cdots \frac{2N}{2N-1}\frac{2N}{2N+1}$$ which looks a bit more like the Wallis product.