Why $\lim_{x\to\infty}\ (\Gamma(x)^{1/x})'=1/e$ I find that the graph of $f(x)=\Gamma(x)^{1/x}$ or $(x!)^{1/x}$ is nearly a line when $x$ is large enough. And when $x$ gets larger, $f'$ will come to a limit.
For example:
x      f'(x)
1    -0.577216
2     0.211392
3     0.290511
4     0.316217
...   ...
10    0.349503
20    0.358720
...   ...
1000  0.367695
2000  0.367787
...   ...
9000  0.367859

(calculated using SymPy)
The answer given by WolframAlpha is $1/e$. It is amazing, and why is it?
PS.
$f'(x)=(\mathrm{polygamma}(0,x)/x-\ln(\Gamma(x))/x^2)*\Gamma(x)^{1/x}$

 A: It is shown in https://doi.org/10.1007/s00013-010-0146-9 that
$$
\Gamma (x) \sim \left( {\frac{x}{e}\left( {1 + \frac{1}{{12x^2 }} +  \cdots } \right)} \right)^x \sqrt {\frac{{2\pi }}{x}} 
$$
for large positive $x$. Combining this with
$$
\sqrt[x]{{\sqrt {\frac{{2\pi }}{x}} }} = \exp \left( {\frac{1}{{2x}}\log \left( {\frac{{2\pi }}{x}} \right)} \right) = 1 + \frac{1}{{2x}}\log \left( {\frac{{2\pi }}{x}} \right) + \frac{1}{{8x^2 }}\log ^2 \left( {\frac{{2\pi }}{x}} \right) +  \cdots 
$$
we find
$$
(\Gamma (x))^{1/x}  \sim \frac{x}{e}\left( {1 + \frac{{\log (2\pi /x)}}{{2x}} + \frac{{3\log ^2 (2\pi /x) + 2}}{{24x^2 }} +  \cdots } \right)
$$
as $x\to +\infty$. You can see that, at leading order,
$$
\frac{d}{{dx}}(\Gamma (x))^{1/x} \sim \frac{d}{{dx}}\frac{x}{e} = \frac{1}{e}.
$$
A: $$y=\Gamma (x)^{\frac{1}{x}}\implies \log(y)=\frac{1}{x}\log (\Gamma (x))$$ Using Stirling approximation
$$\log(y)=\log (x)-1+\frac{\log \left(\frac{2 \pi }{x}\right)}{2 x}+O\left(\frac{1}{x^2}\right)$$
$$y=e^{\log(y)}=\frac{x}{e}+\frac{\log \left(\frac{2 \pi }{x}\right)}{2 e}+O\left(\frac{1}{x}\right)$$
