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

Answer: 
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

$$\log n!=\sum\_{k=2}^n\log k=\int\_1^{n} \log\lceil k\rceil\,dk>\int\_1^{n} \log k\,dk =n\log n-n+1.$$
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 $\lim\limits\_{n\to\infty} \frac{\log(n!)}{n}=\lim\limits\_{n\to\infty} \log(a\_n)= \infty$.