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.
 A: 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 $$(n+1)^{n} \sqrt[n]{n!} \ge \frac{(n+1)^{n+1}}{e} \ge n^{n+1}$$.
A: Taking logarithms the inequality is equivalent to
$$
(n+1)\log n\le n\log(n+1)+\frac1n\sum_{k=1}^n\log k.
$$
We estimate now the sum:
$$
\sum_{k=1}^n\log k\ge\int_1^n\log x\,dx=n\log n-n+1.
$$
It will be enough to prove that
$$
n\log n\le n\log(n+1)-1+\frac1n,
$$
or equivalently
$$
1-\frac1n\le n\log\Bigl(1+\frac1n\Bigr).
$$
This follows from the inequality $\log(1+x)\ge x-x^2/2$, $0<x<1$.
A: $$
\begin{align}
\log\left(n\left(1+\frac1n\right)^{-n}\right)
&=\log(n)-n\log\left(1+\frac1n\right)\\
&=\log(n)+n\log\left(1-\frac1{n+1}\right)\\
&\le\log(n)-\frac{n}{n+1}\\
&\le\log(n)-\frac{n-1}n\\
&=\frac1n\int_1^n\log(x)\,\mathrm{d}x\\
&\le\frac1n\sum_{k=2}^n\log(k)\\
&=\log\left(\sqrt[n]{n!}\right)
\end{align}
$$
Exponentiate and rearrange to get
$$
n^{n+1}\le(n+1)^n\sqrt[n]{n!}
$$
A: If you feel inclined to do "nice" (erm) calculus, you can always choose to study one of these two functions:
$$
f\colon x\in [1,\infty) \mapsto \Gamma(x+1) (x+1)^{x^2} - x^{x^2+x}
$$
or
$$
g\colon x\in [1,\infty) \mapsto \frac{\Gamma(x+1) (x+1)^{x^2}}{x^{x^2+x}}
$$
and show respectively that $f\geq 0$ or $g \leq 1$ on their domain. (for instance, as $g$ is increasing, if I'm not mistaken, so after showing this you would only have to compute $\lim_\infty g$).
This will not be very enjoyable, but should work: whichever of the two functions you choose, it'll be differentiable and nicely behaved. (If you want, you can even define them on $[2,\infty)$ instead if technicalities at $1$ pop up.)
