Monotonicity of $\frac{n}{\sqrt[n]{(n!)}}$

Taking $\log$ gives $$\sum_{i=1}^n\frac{-\log \frac i n}{n}$$ This is a Riemann sum of the integral $\int_{0}^1(-\log x)\,dx$.
Since the function $-\log x$ is decreasing and convex, its Riemann sums have to increase monotonically as the partition gets finer, if the partitions are regular. (Proof here).

We need to show that

$$a_n=\frac{n}{\sqrt[n]{n!}}\quad a_{n+1}\geq a_n\iff$$

$$\frac{n+1}{\sqrt[n+1]{(n+1)!}}\geq \frac{n}{\sqrt[n]{n!}} \iff \frac{n+1}{ (n+1)!^ \left(\frac{1}{n+1}\right) }n!^\left( \frac{1}{n} \right)\geq n \iff$$

$$ (n+1)^{\left(1-\frac{1}{n+1}\right)}n!^{\left(\frac{1}{n} -\frac{1}{n+1}\right)}\geq n \iff (n+1)^{\left(\frac{n}{n+1}\right)}n!^{\left(\frac{1}{n(n+1)}\right)}\geq n $$

$$ (n+1)^{n^2}n!\geq n^{\left(n(n+1)\right)}=n^{n^2}n^n\iff \frac{n^n}{n!}\leq \left(\frac{n+1}{n}\right)^{n^2}=\left(1+\frac1n\right)^{n^2}\iff $$

$$\left(1+\frac1n\right)^{n^2} \geq\frac{n^n}{n!}$$

which is true since

$$\left(1+\frac1n\right)^{n^2}\geq \left(1+\frac1n\right)^{n^2-1}=\left[\left(1+\frac1n\right)^{n+1}\right]^{n-1}\geq e^{n-1}\geq\frac{n^n}{n!}$$

indeed

$$e^{n-1}\geq\frac{n^n}{n!}\iff b_n=\frac {e^nn!}{n^n}\geq e$$

which is true $\forall n$ since

$$n=1 \implies b_1=\frac{e^11!}{1^1}\geq e$$

and

$$\frac{b_{n+1}}{b_n}=\frac {e^{n+1}(n+1)!}{(n+1)^{n+1}}\frac {n^n}{e^nn!}=\frac{e}{\left(1+\frac1n\right)^n}>1$$

For the last inequality see also the related OP

Show that $e^{1-n} \leq \frac {n!}{n^n}$