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For the sake of simplicity let's take $k = 51$, how do I proceed without applying the Sterling's log approximation. I'm looking for a way to find an exact answer. Thank you.

Sambo
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Sanu Kumar
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    What is supposed to be $x!$ ,where $x \in \mathbb{R}/ \mathbb{Z}$ ? – openspace Oct 27 '18 at 21:53
  • @openspace, $x\notin\Bbb Z$ obviously, as $x!$ is even for all $x\ge2, x\in\Bbb Z$, while $51^x$ is most certainly odd. – Rhys Hughes Oct 27 '18 at 21:57
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    @RhysHughes I mean how do we want to determine $x!$ for $x \notin \mathbb{Z}$ – openspace Oct 27 '18 at 22:01
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    Oh I see, different notation. My apologies – Rhys Hughes Oct 27 '18 at 22:04
  • @openspace there is a way to define the factorial for real numbers, through an integral. – RGS Oct 27 '18 at 22:18
  • If $x$ and $k$ are natural numbers that satisfy $k^x = x!$, then $(x,k) = (0,0),,(0,1),,(1,1)$. If $k$ is a real number, $k = (x!)^{1/x}$. If $x$ is a real number, you're going to need to specify which extension of $x!$ to the reals you want to use. – eyeballfrog Oct 27 '18 at 22:22
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    @RGS that's would be gamma function, am I right? I mean that's not the definition of factorial. Gamma function is more general. – openspace Oct 27 '18 at 22:29

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