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),
$$\lim\limits\_{n\to\infty} \frac{\log(n!)}{n}=\lim\limits\_{n\to\infty} \frac{\sum\_{k=1}^n\log{k}}{n}=\lim\limits\_{n\to\infty} \frac{\log (n+1)}{(n+1)-n}.$$
The expression is the logarithm of

$$\frac1e\cdot\frac2e\cdot\frac3e\cdots\frac ne$$ which obviously diverges.

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You can also use

$$\log n!=\sum\_{k=2}^n\log k=\int\_1^{n} \log\lceil k\rceil\,dk>\int\_1^{n} \log k\,dk =n\log n-n+1.$$