TITLE: Prove that $n^{n+1} \leq (n+1)^{n} \sqrt[n]{n!}$
QUESTION [18 upvotes]: Let $n$ be a positive integer. I conjectured that the following inequality is true
\begin{equation}
n^{n+1} \leq (n+1)^{n} \sqrt[n]{n!} .
\end{equation}
Anyhow I could neither prove nor disprove it. I could only check, by using Stirling's Formula, that the ratio of the right and left members tends to 1 as $n \rightarrow \infty$. Any help is welcome.

REPLY [17 votes]: $$
\begin{align}
\log\left(n\left(1+\frac1n\right)^{-n}\right)
&=\log(n)-n\log\left(1+\frac1n\right)\\
&=\log(n)+n\log\left(1-\frac1{n+1}\right)\\
&\le\log(n)-\frac{n}{n+1}\\
&\le\log(n)-\frac{n-1}n\\
&=\frac1n\int_1^n\log(x)\,\mathrm{d}x\\
&\le\frac1n\sum_{k=2}^n\log(k)\\
&=\log\left(\sqrt[n]{n!}\right)
\end{align}
$$
Exponentiate and rearrange to get
$$
n^{n+1}\le(n+1)^n\sqrt[n]{n!}
$$