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

Answer: 
$$\log\operatorname{GM}(1,\ldots,n)\geq \log\operatorname{HM}(1,\ldots,n)=\log\left(\frac{n}{H\_n}\right)\geq \log n-\log\left(\log(n+1)+1\right). $$
$\text{GM}$: geometric mean. $\text{HM}$: harmonic mean. $H\_n$: harmonic number, $\sum\_{k=1}^{n}\frac{1}{k}\leq 1+\int\_{1}^{n+1}\frac{dx}{x}$.
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 $\lim\limits\_{n\to\infty} \frac{\log(n!)}{n}=\lim\limits\_{n\to\infty} \log(a\_n)= \infty$.