Is $(n+1)! = \mathrm{o}(n^n)$? Continuing http://math.stackexchange.com/a/2431263, it seems to me that you can prove with the same method that $(n+1)! = o(n^n)$. 
Namely:
Let $a_n=\frac{(n+1)!}{n^n}$. Then:
$$\frac{a_{n+1}}{a_n}=\frac{\frac{(n+2)!}{(n+1)^{n+1}}}{\frac{(n+1)!}{n^n}}=\frac{n^n(n+2)!}{(n+1)!(n+1)^{n+1}}=\frac{n^n}{(n+1)^n}\frac{n+2}{n+1}=\left(\frac{n}{n+1}\right)^n \left(1+\frac{1}{n+1}\right) \to \frac{1}{e}\cdot 1 <1$$
Thus from ratio test for sequences we have that $a_n \to 0$.
Can anyone confirm or reject this?
 A: Compare
$$LHS=\color{green}2\cdot3\cdot4\cdots n\cdot\color{green}{(n+1)}$$ and $$RHS=\color{green}n\cdot n\cdot n\cdot n\cdots \color{green}n$$ ($n$ factors each).
Now it is clear that $LHS=o( RHS)$ just because $\color{green}2\cdot\color{green}{(n+1)}=o(\color{green}{n\cdot n})$ and all intermediate factors are smaller or equal.

More generally, for any constant $k$, $(n+k)!=o(n^n)$.
A: Your answer is close, but it seems more satisfying to use $\frac{n+2}{n}\frac1{\left(1+\frac1n\right)^{n+1}}$ as the ratio because it gives telescoping inequalities.

Let
$$
a_n=\frac{(n+1)!}{n^n}\tag1
$$
then, because $\frac{n+2}{n}\le\left(\frac{n+1}{n}\right)^2$ (expand the square) and $\left(1+\frac1n\right)^{n+1}\ge e$ (see this answer, which says that $\left(1+\frac1n\right)^{n+1}$ decreases to $e$),
$$
\begin{align}
\frac{a_{n+1}}{a_n}
&=(n+2)\frac{n^n}{(n+1)^{n+1}}\\
&=\frac{n+2}{n}\frac1{\left(1+\frac1n\right)^{n+1}}\\
&\le\left(\frac{n+1}{n}\right)^2\frac1e\tag2
\end{align}
$$
Therefore, telescopically, we have
$$
a_n\le a_1\frac{n^2}{e^{n-1}}\tag3
$$
and since
$$
\lim_{n\to\infty}\frac{n^2}{e^{n-1}}=0\tag4
$$
we get that $(n+1)!=o\!\left(n^n\right)$
