Is this conjecture correct ${(2\pi)^{m-1}\Gamma(m)\over m^m}=\prod_{k=1}^{\infty}{mk\over mk-(m-1)}\cdot{mk\over mk+(m-1)}?$

Question

The Wallis's Product

$${\pi\over 2}={2\over 1}\cdot{2\over 3}\cdot{4\over 3}\cdot{4\over 5}\cdot{6\over 5}\cdot{6\over 7}\cdots\tag1$$

Conjecture: The generalised of $(1)$

$${(2\pi)^{m-1}\Gamma(m)\over m^m}=\prod_{k=1}^{\infty}{mk\over mk-(m-1)}\cdot{mk\over mk+(m-1)}\tag2$$

$m\ge2$ and $\Gamma(m)$ is the Gamma function

How do we prove $(2)?$

Answer 1

Note we have the following infinite product: $$\frac{\sin x}{x} = \prod_{n=1}^\infty\left(1 - \frac{x^2}{n^2\pi^2}\right)$$ Putting $x=\dfrac{(m-1)\pi}{m}$ we have $$\dfrac{m}{(m-1)\pi}\sin \left(\frac{(m-1)\pi}{m} \right) = \prod_{n=1}^\infty\left(1 - \frac{(m-1)^2}{m^2n^2}\right)$$ Taking the reciprocal we have $$\frac{(m-1)\pi}{m} \csc \left(\frac{(m-1)\pi}{m} \right) =\prod_{n=1}^{\infty}{mn\over mn-(m-1)}\cdot{mn\over mn+(m-1)}$$ Now $$\frac{(m-1)\pi}{m} \csc \left(\frac{\pi}{m} \right) =\prod_{n=1}^{\infty}{mn\over mn-(m-1)}\cdot{mn\over mn+(m-1)}$$ So your conjecture becomes $$\csc \left(\frac{\pi}{m} \right)=\frac{(2\pi)^{m-1} \Gamma(m)}{m^{m-1}(m-1)\pi}$$ Which is wrong, as can be seen if you put $m=4$.