Monotonicity of $\frac{n}{\sqrt[n]{(n!)}}$ It is known that when $n\rightarrow\infty$ the sequence $$\frac{n}{\sqrt[n]{(n!)}}$$ has limit $e$ but I don't know how to prove it's monotonicity. After a short calculus using WolframAlpha I found that this sequence is actually increasing. I tried to compare $2$ consecutive members but I couldn't manage to show something. It is obvious that $$\frac{n^n}{n!}$$ is increasing but that don't help us much (I think)

The ratio of the consecutive terms of the sequence $a_n=\frac{n}{\sqrt[n]{n!}}$ is greater than $1$.
Therefore, $\frac{n}{\sqrt[n]{n!}}$ is increasing.
$$
\begin{align}
\left(\frac{a_{n+1}}{a_n}\right)^{n(n+1)}
&=\left[\frac{\frac{\color{#C00}{n+1}}{\color{#090}{\sqrt[n+1]{(n+1)!}}}}{\frac{\color{#C00}{n}}{\color{#090}{\sqrt[n]{n!}}}}\right]^{n(n+1)}\tag1\\
&=\left(\color{#C00}{\frac{n+1}n}\right)^{n(n+1)}\color{#090}{\frac{n!^{n+1}}{(n+1)!^n}}\tag2\\[6pt]
&=\left(\frac{n+1}n\right)^{n(n+1)}\frac{n!}{(n+1)^n}\tag3\\[6pt]
&=\left(\frac{n+1}n\right)^{n^2}\frac{n!}{n^n}\tag4\\[15pt]
&\ge2\tag5
\end{align}
$$
Explanation:
$(2)$: distribute the exponent $n(n+1)$ over the ratios
$(3)$: cancel $n!^n$ in the numerator and denominator of the right term
$(4)$: move $\left(\frac{n+1}n\right)^n$ from the left term to the right term
$(5)$: inequality $(10)$ shows that $b_n=\left(\frac{n+1}n\right)^{n^2}\frac{n!}{n^n}$ is increasing
$\phantom{\text{(5):}}$ since $b_1=2$, $b_n\ge2$

Bernoulli's Inequality shows that $b_n=\left(\frac{n+1}n\right)^{n^2}\frac{n!}{n^n}$ is increasing:
$$
\begin{align}
\frac{b_{n+1}}{b_n}
&=\frac{\color{#C00}{\left(\frac{n+2}{n+1}\right)^{(n+1)^2}}\color{#090}{\frac{(n+1)!}{(n+1)^{n+1}}}}{\color{#C00}{\left(\frac{n+1}n\right)^{n^2}}\color{#090}{\frac{n!}{n^n}}}\tag6\\
&=\color{#C00}{\left(\frac{n(n+2)}{(n+1)^2}\right)^{(n+1)^2}\left(\frac{n+1}n\right)^{2n+1}}\color{#090}{\left(\frac{n}{n+1}\right)^n}\tag7\\
&=\left(1-\frac1{(n+1)^2}\right)^{(n+1)^2}\left(\frac{n+1}n\right)^{n+1}\tag8\\[6pt]
&\ge\left(1-\frac1{n+1}\right)^{n+1}\left(\frac{n+1}n\right)^{n+1}\tag9\\[12pt]
&=1\tag{10}
\end{align}
$$
Explanation:
$\phantom{1}(7)$: multiply numerator and denominator by $\left(\frac{n+1}n\right)^{2n+1}$ and combine terms
$\phantom{1}(8)$: $\frac{n(n+2)}{(n+1)^2}=1-\frac1{(n+1)^2}$
$\phantom{1}(9)$: Bernoulli's Inequality
$(10)$: $1-\frac1{n+1}=\frac{n}{n+1}$
