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

enter image description here

but fails on this one

enter image description here

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?

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  • $\begingroup$ 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. $\endgroup$ – mikado Nov 18 at 20:04
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    $\begingroup$ I've confirmed this behaviour in V12.0. It would be interesting to know if earlier versions give the same result. $\endgroup$ – mikado Nov 18 at 20:04
  • $\begingroup$ Also fails on Mac OS 11.3 $\endgroup$ – Rohit Namjoshi Nov 18 at 20:27
  • $\begingroup$ @mikado: Also fails on 11.3.0 for Linux ARM (32-bit) $\endgroup$ – g.kov Nov 22 at 6:16
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Workaround:

$$\Gamma \left(\frac{k}{n}\right)=\frac{\Gamma \left(\frac{k}{n}+1\right)}{\frac{k}{n}}$$

$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] *)
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  • $\begingroup$ How would you display it as shown in the question? //TraditionalForm does not. $\endgroup$ – nilo de roock Nov 19 at 8:33
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    $\begingroup$ @nilo, (2 π)^(1/2 (-1 + n))/Sqrt[n] // (TraditionalForm[Sqrt[#^2]] &) at least works for this case. $\endgroup$ – J. M. will be back soon Nov 19 at 8:35
  • $\begingroup$ Wow! Did not know that was possible. So cool. $\endgroup$ – nilo de roock Nov 19 at 8:40
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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
(* "12.0.0 for Linux x86 (64-bit) (April 15, 2019)" *)

Limit[Product[Gamma[z + k/n], {k, 1, n - 1}], z -> 0]
(* (2 \[Pi])^(1/2 (-1 + n))/Sqrt[n] *)
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