Use the fact that $\Gamma (1-z) = -z\, \Gamma(-z)$ and then:
$$\Gamma(1-z)\Gamma(z) = -z \, \Gamma(-z)\Gamma(z) = -z \cdot \frac{1}{-z}\cdot \frac{1}{z} \prod_{k=1}^{+\infty} \frac{1}{\left(1 + \frac{z}{k} \right)\left(1 - \frac{z}{k} \right) } = \frac{1}{z} \prod_{k=1}^{+\infty} \frac{1}{1 - \frac{z^2}{k^2}} = \frac{\pi}{\sin \pi z}$$
There are several proofs for the sine product, e.g. look for “sine product formula proof”.
You can also take a look here, there is a link to the old proof of Euler.
For a proof of the Wallis product it’s enough to look e.g. here.
It’s $~\displaystyle \frac{\pi}{2} = \frac{\pi x}{\sin(\pi x)}|_{x=\frac{1}{2}} = \Gamma(1-x)\Gamma(1+x)|_{x=\frac{1}{2}} = \prod\limits_{n=1}^\infty \frac{2n}{2n-1}\frac{2n}{2n+1}~$ .
The connection to the Riemann zeta function can be seen by using the logarithm.
$\displaystyle \ln\frac{\pi x}{\sin(\pi x)} = \ln(\Gamma(1-x)\Gamma(1+x)) = \sum\limits_{n=1}^\infty\frac{x^{2n}}{n}\zeta(2n)$
The last equation comes from $~\displaystyle \ln\prod\limits_{k=1}^n (1-zb_k)^{a_k} = \sum\limits_{k=1}^\infty\frac{z^k}{k}\sum\limits_{v=1}^n a_v b_v^k$
with $~\displaystyle (a_v;b_v;n;z):=\left(1;\frac{1}{v^2};\infty;x^2\right)~$ .
To get the formula for $~\zeta(2n)~$ one should use the Bernoulli numbers $\,B_k\,$,
traditionelly introduced by $\displaystyle \frac{x}{e^x-1} = \sum\limits_{k=0}^\infty \frac{x^k}{k!}B_k$ .
With the first derivation of $~\displaystyle \ln\frac{\pi x}{\sin(\pi x)}~$ follows:
$\displaystyle \frac{1 -\pi x\cot(\pi x)}{x} =\frac{d}{dx}\ln\frac{\pi x}{\sin(\pi x)} = \frac{d}{dx}\sum\limits_{n=1}^\infty\frac{x^{2n}}{n}\zeta(2n) = 2\sum\limits_{n=1}^\infty x^{2n-1}\zeta(2n)$
with $\enspace\displaystyle \cot(z) = \frac{\cos z}{\sin z} = i\frac{ e^{iz}+e^{-iz} }{ e^{iz}-e^{-iz} } = i\frac{ e^{iz}-e^{-iz}+2e^{-iz} }{ e^{iz}-e^{-iz} } = i\left(1+\frac{1}{iz}\frac{i2z}{e^{i2z}-1}\right)$
$\hspace{2.3cm}\displaystyle = i\left(1+\frac{1}{iz}\sum\limits_{k=0}^\infty \frac{(i2z)^k}{k!}B_k\right) = \frac{1}{z}+\frac{1}{z}\sum\limits_{k=1}^\infty (-1)^k\frac{(2z)^{2k}}{(2k)!}B_{2k}$
We get $\displaystyle \enspace\frac{1 -\pi x\cot(\pi x)}{x} = \sum\limits_{k=1}^\infty (-1)^{k-1}\frac{(2\pi)^{2k}x^{2k-1}}{(2k)!}B_{2k}$
and comparing the coefficients of $~x^{2k-1}~$ it follows $~\displaystyle \zeta(2n)=(-1)^{k-1}\frac{(2\pi)^{2k}}{2(2k)!}B_{2k}~$ .
Best Answer
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.