Evaluate the limit: $$ \lim_{n\to\infty}\frac{\log_a n!}{n^b}, \ \ n\in\Bbb N $$
I've tried to consider two cases: $b < 0$, $b \ge 0$. First $b < 0$. This case is simple since the limit becomes: $$ \lim_{n\to\infty}\frac{\log_a n!}{n^b} = \lim_{n\to\infty}n^{|b|}\log_a n! = +\infty $$
Now consider the case when $b \ge 0$, then we may apply Cesaro-Stolz theorem, then the limit is equal to the following limit: $$ \begin{align} \lim_{n\to\infty} \frac{\log_a n!}{n^b} &= \lim_{n\to\infty} \frac{\log_a (n+1)! - \log_an!}{(n+1)^b - n^b} \\ &= \lim_{n\to\infty} \frac{\log_a(n+1)}{(n+1)^b - n^b} \end{align} $$
I've shown earlier that: $$ \lim_{n\to\infty}\left((n+1)^b - n^b\right) = 0,\ \text{if}\ b\in(0, 1)\\ \lim_{n\to\infty}\left((n+1)^b - n^b\right) = +\infty,\ \text{if}\ b > 1\\ $$
For $b = 1$ the limit becomes: $$ \lim_{n\to\infty}{\log_an!\over n} = \lim_{n\to\infty}\log_a\sqrt[n]{n!} = +\infty $$
So it looks like: $$ b \le 1 \implies \lim_{n\to\infty}\frac{\log_a n!}{n^b} = +\infty $$
Here I'm not sure how to handle the case for $b > 1$. What are the steps to handle the case for $b > 1$? I know the limit is $0$, but want to justify that.