Q:

Proving $n!n^s=o(n^n)$

I want to prove that for any $s\geq0$, and $n\to\infty$, $n!n^s=o(n^n)$ or $$\lim_\limits{n\to\infty}\frac{n!n^s}{n^n}=0.$$ As a hint, the inequality $$\sum\limits_{k=1}^{n}\log k\leq n\log\frac{n+1}2$$ is given.
I thought that instead of proving $\lim_\limits{n\to\infty}n!n^{s-n}=0$, one could also show $\lim_\limits{n\to\infty}\log\left(n!n^{s-n}\right)=-\infty$. This would be equivalent to $\lim_\limits{n\to\infty}\log\left(n!\right)+\log\left(n^{s-n}\right)=-\infty$. Using the given inequality, it would be sufficient to prove $$\lim_\limits{n\to\infty}n\log\frac{n+1}2+\log\left(n^{s-n}\right)=-\infty\Leftrightarrow\lim_\limits{n\to\infty}n\log\frac{n+1}{2n}+s\log n=-\infty\\\Leftrightarrow\lim_\limits{n\to\infty}n\log\frac12+s\log n=-\infty.$$
What should I do now?

A:

The series
$$\sum_\limits{n=1}^\infty\frac{n!n^s}{n^n}$$
is convergent by the ratio test (easy exercise). Therefore
$$\lim_\limits{n\to\infty}\frac{n!n^s}{n^n}=0.$$

