<s>[INST] Show that $\frac{(m!)^{1/m}}{m/e}$ is a decreasing function of $m$ Show that
$\dfrac{(m!)^{1/m}}{m/e}$
is a decreasing function of $m$.
Here is my proof.
I would like to see others,
preferably simpler.
I have shown in 
Proof explanation $\lim\limits_{n\to\infty}\frac{n}{\sqrt[n]{n!}}=e$
that,
if
$r_m
=\dfrac{(m!)^{1/m}}{m/e}
$,
then
$r_m > 1+1/m
$.
This was based on
$(1+1/m)^m < e
\lt (1+1/m)^{m+1}
$,
which was also shown there.
Since
$r_m \to 1$,
this leads to the conjecture that
$r_m$ is a decreasing function of $m$.
Let
$s_m = \dfrac{r_{m+1}}{r_m}$.
Then,
since $r_m > 1+1/m$
implies that
$m! > (1+1/m)^mm^m/e^m$,
$\begin{array}\\
s_m 
&= \dfrac{r_{m+1}}{r_m}\\
&= \dfrac{\dfrac{((m+1)!)^{1/(m+1)}}{(m+1)/e}}{\dfrac{(m!)^{1/m}}{m/e}}\\
&= \dfrac{\dfrac{((m+1)!)^{1/(m+1)}}{(m+1)}}{\dfrac{(m!)^{1/m}}{m}}\\
&= \dfrac{m((m+1)!)^{1/(m+1)}}{(m+1)(m!)^{1/m}}\\
\text{so}\\
s_m^{m(m+1)} 
&= \dfrac{m^{m(m+1)}((m+1)!)^{m}}{(m+1)^{m(m+1)}(m!)^{m+1}}\\
&= \dfrac{m^{m(m+1)}m!^m(m+1)^m}{(m+1)^{m(m+1)}m!^{m}m!}\\
&= \dfrac{m^{m(m+1)}}{(m+1)^{m^2}m!}\\
&= \dfrac{m^{m}}{(1+1/m)^{m^2}m!}\\
&< \dfrac{m^{m}}{(1+1/m)^{m^2}(1+1/m)^mm^m/e^m}\\
&= \dfrac{e^m}{(1+1/m)^{m^2+m}}\\
\text{so}\\
s_m^{m+1} 
&< \dfrac{e}{(1+1/m)^{m+1}}\\
&< 1\\
\end{array}
$
I find it pleasing that
the proof ends up
depending on the
upper bound for $e$.
 [/INST] I guess the following is "quicker", but relies on some heavy machinery. We have
\begin{align*}
\frac{d}{dm} \log r_m &= -\frac{1}{m^2} \log m! + \frac{1}{m} \frac{\Gamma'(m+1)}{m!} - \frac{1}{m} \\
&\le -\frac{\log\sqrt{2\pi}-m+(m+\frac{1}{2})\log m}{m^2} + \frac{H_m - \gamma - 1}{m}
\end{align*}
where we used the fact that $\sqrt{2\pi}m^{m+\frac{1}{2}}e^{-m} \le m!$ and $\Gamma'(m+1) = m!(-\gamma + H_m)$, where $H_m$ is the $m$th harmonic number. So
\begin{align*}
\frac{d}{dm} \log r_m &\le \frac{1}{m}\left(H_m - \left(1 + \frac{1}{2m}\right)\log m - \gamma - \frac{1}{m}\log\sqrt{2\pi}\right)
\end{align*}
But since
\begin{align*}
H_m &= \log m + \gamma + \frac{1}{2m} - \frac{1}{12m^2} + \cdots \\
&< \log m + \gamma + \frac{1}{2m}  \\
&< \left(1 + \frac{1}{2m}\right)\log m + \gamma + \frac{1}{m}\log\sqrt{2\pi}
\end{align*}
we are done.
</s>