<s>[INST] Approximating factorial using identity $\frac{1}{x}!\frac{2}{x}!\cdots\cdot\frac{x}{x}!=\frac{ {x}!\cdot(2\pi)^{\frac{x-1}{2}} }{ x^x\cdot\sqrt{x} }$ I created a function that describes the product of the inverse multiples of a factorial
$$ m(x) = \frac{1}{x}!\cdot\frac{2}{x}!\cdot\frac{3}{x}!\cdots\frac{x-1}{x}!\cdot\frac{x}{x}!$$
for some reasons i thought this function might be useful, thats why i'm posting an incomplete explanation to the best of my knowledge but after some rigorous calculation, i was able to express $m(x)$ as a formula
$$ m(x) = \frac{ {x}!\cdot(2\pi)^{\frac{x-1}{2}} }{ x^x\cdot\sqrt{x} }$$
it turns out that the graph of $m(x)$ is a very simple one, makes me suggest it would be easy to interpolate and create a super factorial approximation from it
$$ {x}! = x^x\cdot\sqrt{x}\cdot\sqrt{2\pi}\cdot m(x)\cdot\sqrt{{2\pi}^{-x}}$$
and it resembles striling's approximation
$${x}! \approx x^x\cdot\sqrt{x}\cdot\sqrt{2\pi}\cdot e^{-x} $$

$$
m(1)=1, m(2)=\frac{1}{2}!, m(3)=\frac{4\pi}{3^{5/2}}, m(4)=\frac{3\pi^{3/2}}{2^{9/2}}, m(5)=\frac{96\pi^2}{5^{9/2}}, m(6)=\frac{45\pi^{5/2}}{3^{13/2}} , m(7)=\cdots
$$
now my question, can you help with the approximation, does the function gives any more information
 [/INST] The Gauss multiplication formula states that
$$
\prod\limits_{k = 0}^{n - 1} {\Gamma\! \left( {z + \frac{k}{n}} \right)}  = (2\pi )^{(n - 1)/2} n^{1/2 - nz} \Gamma (nz)
$$
for $n\geq 1$ and for all complex $z$ for which both sides are defined. Taking $z=1$ and using $r!=\Gamma(1+r)$ gives
$$
\prod\limits_{k = 1}^n {\left( {\frac{k}{n}} \right)!}  = \prod\limits_{k = 0}^{n - 1} {\left( {\frac{k}{n}} \right)!}  = \frac{{(2\pi )^{(n - 1)/2} n!}}{{n^n \sqrt n }}.
$$
Thus your result is a special case of the Gauss multiplication formula. An asymptotic expansion for $m$ coming from the Stirling series is
$$
m(x) \sim \left( {\frac{{\sqrt {2\pi } }}{e}} \right)^x \left( {1 + \frac{1}{{12x}} + \frac{1}{{288x^2 }} - \frac{{139}}{{51840x^3 }} -  \cdots } \right)
$$
as $x\to +\infty$.
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