Proof of $\frac{e}{(n+1)^{\frac{1}{n}}} \frac{n}{n+1} < \frac{n}{n!^{\frac{1}{n}}}$ Original Question:

Use the proof of Thm 7.2.4 given above to show that if $n \geq 2$, then $\frac{e}{(n+1)^{\frac{1}{n}}} \frac{n}{n+1} < \frac{n}{n!^{\frac{1}{n}}} < \frac{e}{4^{\frac{1}{n}}}$.


Thm 7.2.4: $\displaystyle\lim_{n\rightarrow \infty} \dfrac{n}{(n!)^\frac{1}{n}} = e$


Proof: Let $a_n = \dfrac{n^n}{n!}$, i.e. $\dfrac{a_{n+1}}{a_n} = \left(\dfrac{n+1}{n}\right)^n \rightarrow e$

Source: Limits - A New Apporach to Real Analysis - Alan F. Beadon - Undergraduate Texts in Mathematics - Springer
In particular, I have already proved the 2nd inequality. However, IDK how to solve the first inequality.
I can show that $\frac{(n+1)^{(n+1)}}{n!}>e^n$, but I don't know the further steps. Any hint? Thanks.

You can use induction. Check the case $n=2$ by hand. Then
\begin{align*}
\frac{{(n + 1)^{n + 1} }}{{(n + 1)!}} & = \frac{{n^n }}{{n!}}\frac{{(n + 1)^n }}{{n^n }} > \left( {\frac{n}{{n + 1}}} \right)^n \frac{{e^n }}{{n + 1}}\frac{{(n + 1)^n }}{{n^n }} = \frac{{e^n }}{{n + 1}} \\ &= \left( {\frac{{n + 1}}{{n + 2}}} \right)^{n + 1} \frac{{e^{n + 1} }}{{n + 2}}\frac{1}{e}\left( {1 + \frac{1}{{n + 1}}} \right)^{n + 2}  > \left( {\frac{{n + 1}}{{n + 2}}} \right)^{n + 1} \frac{{e^{n + 1} }}{{n + 2}}.
\end{align*}
In the last step we used $e < \left( {1 + \frac{1}{m}} \right)^{m + 1}$.
