Closed form of product of Gamma function

Mathematica recognizes this closed form \begin{align} \prod_{k=1}^{n-1}\sin(\pi k/n) &= 2^{1-n}\,n \end{align} just fine:

but fails on this one

despite that this expression also has a known closed form \begin{align} \prod_{k=1}^{n-1}\Gamma(k/n) &= \sqrt{\frac{ (2\,\pi)^{n-1}}{n}} . \end{align}

Is there a way to make Mathematica to recognize it?

• This is peculiar, since for a finite product, there is no obvious reason why it should need to consider any evaluation with a non-finite result. – mikado Nov 18 '19 at 20:04
• I've confirmed this behaviour in V12.0. It would be interesting to know if earlier versions give the same result. – mikado Nov 18 '19 at 20:04
• Also fails on Mac OS 11.3 – Rohit Namjoshi Nov 18 '19 at 20:27
• @mikado: Also fails on 11.3.0 for Linux ARM (32-bit) – g.kov Nov 22 '19 at 6:16

$$\Gamma \left(\frac{k}{n}\right)=\frac{\Gamma \left(\frac{k}{n}+1\right)}{\frac{k}{n}}$$
$Version (* "12.0.0 for Microsoft Windows (64-bit) (April 6, 2019)" *) Product[Gamma[k/n + 1]/(k/n), {k, 1, n - 1}] (* (2 \[Pi])^(1/2 (-1 + n))/Sqrt[n] *)  • How would you display it as shown in the question? //TraditionalForm does not. – nilo de roock Nov 19 '19 at 8:33 • @nilo, (2 π)^(1/2 (-1 + n))/Sqrt[n] // (TraditionalForm[Sqrt[#^2]] &) at least works for this case. – J. M. will be back soon Nov 19 '19 at 8:35 • Wow! Did not know that was possible. So cool. – nilo de roock Nov 19 '19 at 8:40 The indeterminate can be overcome using the full identity for $$\Gamma(nz)$$: $$\Gamma(nz)=(2\pi)^{(1-n)/2}n^{nz-1/2}\prod_{k=0}^{n-1}\Gamma(z+\frac{k}{n})$$ and taking the limit as $$z\rightarrow 0$$: $Version