<s>[INST] Factorial identity $\left(\tfrac{1}{2}\right)!$ to get Waallis I asked the wrong question here, my fault :(
How does one see, using $n! = \prod_{k=1}^\infty \left(\frac{k+1}{k}\right)^n \frac{k}{k+n}$, that
$$\left(\frac{1}{2}\right)! = \prod_{k=1}^\infty \left(\frac{k+1}{k}\right)^{\tfrac{1}{2}}\left( \frac{k}{k+\tfrac{1}{2}}\right) = \tfrac{1}{2}\sqrt{2 \cdot \left(\frac{2 \cdot 2}{1 \cdot 3}\frac{4 \cdot 4}{3 \cdot 5}\dotsm \right)}? $$
In other words, derive 
$$ \prod_{k=1}^{\infty}\left(\frac{(2k)^2}{(2k+1)(2k-1)}\right)$$
while manipulating 
$$\prod_{k=1}^\infty \left(\frac{k+1}{k}\right)^{\tfrac{1}{2}}\left( \frac{k}{k+\tfrac{1}{2}}\right)$$
Note this is a way to 'derive' the Wallis product
$$ \tfrac{1}{2}\sqrt{2 \cdot \left(\frac{2 \cdot 2}{1 \cdot 3}\frac{4 \cdot 4}{3 \cdot 5}\dotsm \right)} = \frac{\sqrt{2 \tfrac{\pi}{2}}}{2} = \frac{\sqrt{ \pi}}{2}$$
It should follow, "after a bit of manipulation", but it just doesn't! I must be missing something!
Edit: I now think it is not possible directly!
 [/INST] Looks like you need to show that
$$\prod_{k=1}^{\infty}\left(\frac{(2k)^2}{(2k+1)(2k-1)}\right) = \frac{\pi}{2}.$$
This Wikipedia article uses Euler's infinite product for the sine function:
$$\frac{\sin x}{x} = \prod_{k=1}^{\infty}\left(1 - \frac{x^2}{k^2 \pi^2}\right).$$
The result follows quickly by applying $x = \pi/2.$
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