Question: I am not to find an upper bound to this succession. A lower bound seem to be $\log(n)$. Any ideas?

Answer: 
Hint. By the [Stolz-Cesàro Theorem](https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem),
$$\lim\limits\_{n\to\infty} \frac{\log(n!)}{n}=\lim\limits\_{n\to\infty} \frac{\sum\_{k=1}^n\log{k}}{n}=\lim\limits\_{n\to\infty} \frac{\log (n+1)}{(n+1)-n}.$$
Let $a\_n= (n!)^{1/n}$. Since the power series $\sum\_{n=0}^{\infty}\frac{x^n}{n!}$ has radius of convergence $ \infty$, we have

$a\_n \to \infty$.

Then $\lim\limits\_{n\to\infty} \frac{\log(n!)}{n}=\lim\limits\_{n\to\infty} \log(a\_n)= \infty$.