Prove that $n^{n+1} \leq (n+1)^{n} \sqrt[n]{n!}$ 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.

From $$
e^{-1} = \left(e^{-\frac{1}{n+1}}\right)^{n+1} > \left(1-\frac{1}{n+1}\right)^{n+1}(\because e^x \ge x+1)
$$
and
$$
\left(1-\frac{1}{n+1}\right)^{n+1} \cdot \left(1 + \frac{1}{n}\right)^{n+1} = 1
$$
Note
$$e \le (1+\frac{1}{n})^{n+1}=\frac{(1+n)^{n+1}}{n^{n+1}} \Leftrightarrow n^{n+1} \le \frac{(n+1)^{n+1}}{e} $$
Also note
$$n!e^n=\sum_{k=0}^{\infty}{n^k\frac{n!}{k!}} \ge \sum_{k=0}^{n}{n^k\frac{n!}{k!}} \ge \sum_{k=0}^{n}{n^k\binom{n}{k}}= ({n+1})^n \Leftrightarrow \sqrt[n]{n!} \ge \frac{n+1}{e} $$
This gives us that $$(n+1)^{n} \sqrt[n]{n!} \ge \frac{(n+1)^{n+1}}{e} \ge n^{n+1}$$.
