As stated in the title I am trying to find a good approximation in O notation for: $$(\log n)^n = \log (n) \cdot \log (n) \cdot ... \cdot \log (n)$$. I have tried using L'Hôpital's rule for the term $ \lim_{x\to \infty} \frac{(\log n)^n}{2^n}$ to get a starting point. But I don't think that it helps much in this case as the derivative of $(\log n)^n$ is something along the lines of $\ln^x\left(x\right)\left(\ln\left(\ln\left(x\right)\right)+\dfrac{1}{\ln\left(x\right)}\right)$ and does not really simplify the term.
(How can we get a good approximation for the term?)
EDIT: Is there a closer bound than the trivial $(\log n)^n \in O(n^n)$. And if there is how do we prove that? EDIT2: Closest lower bound I have found is: $(\log n)^n \in \Omega(a^n)$ where $a \in \mathbb{N}^+$. Since $ \lim_{x\to \infty} \frac{(\log n)^n}{a^n} = \lim_{x\to \infty} (\frac{\log n}{a})^n = \infty$, $\log n$ is bigger and therefore $(\log n)^n \in \Omega(a^n)$.
I'm writing this as an answer as it's too large to fit into a comment. The $n!$ bound follows from Stirling's formula which states
$$\log(n!) = n\log{n}-n+O(\log{n}).$$
As I said in the comment $(\log{n})^n = n^{n\frac{\log\log{n}}{\log{n}}}$, and using Stirling's formula gives
$$n! = n^{\frac{log(n!)}{\log{n}}} = n^{\frac{1}{\log{n}}(n\log{n}-n+O(\log{n}))} = n^{n\left(1-\frac{1}{\log{n}} + O(1/n)\right)}.$$
Looking at the exponents, you can see that $n!$ is larger, but the exponent is still asymptotic to $n$ which is why I think $(\log{n})^n = n^{o(n)}$ is better, since $\frac{\log\log{n}}{\log{n}}$ approaches zero as $n\to\infty$. Also, this translates to an upper bound of the form $n^{\epsilon\cdot n}$ for any $\epsilon \gt 0$, which beats $n!$ for any $\epsilon \lt 1$.