TITLE: A question about inequality ${(n+1)\over e^n}^n<n!$
QUESTION [5 upvotes]: How to prove the inequality $${(n+1)\over e^n}^n<n!$$
I have tried mathematical induction, but it doesn't work! Are there other methods to solve it?

REPLY [3 votes]: Let
$$
a_n=\frac{(n+1)^n}{n!}
$$
then
$$
\begin{align}
a_n
&=\left(1+\frac1n\right)^n\frac{n^n}{n!}\\
&=\left(1+\frac1n\right)^n\frac{n^{n-1}}{(n-1)!}\\
&=\left(1+\frac1n\right)^n\,a_{n-1}\\[9pt]
&\le ea_{n-1}
\end{align}
$$
Since $a_1=2\lt e$, inductively, we have that $a_n\lt e^n$.