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

Answer: 
We have:

$$n! = n\cdot (n-1) \cdots 2 \cdot 1 > \underbrace{\frac{n}2 \cdot \frac{n}2 \cdot \cdots \cdot \frac{n}2}\_{\frac{n}2 \text{ times}} \underbrace{\cdot 1 \cdot \cdots \cdot 1}\_{\frac{n}2 \text{ times}} = \left(\frac{n}2\right)^{\frac{n}2}$$

$$n! = n\cdot (n-1) \cdots 2 \cdot 1 < n \cdot n \cdot \cdots \cdot n = n^n$$

So:

$$\frac12 \cdot \log \frac{n}2= \frac{n}2 \cdot \frac{\log \frac{n}2}{n}= \frac{\log \left(\frac{n}2\right)^{\frac{n}2}}{n} <\frac{\log n!}{n} < \frac{\log n^n}{n} = \frac{n\log n}{n} = \log n$$

This means the sequence diverges to $+\infty$.
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 $\lim\limits\_{n\to\infty} \frac{\log(n!)}{n}=\lim\limits\_{n\to\infty} \log(a\_n)= \infty$.