I want to show, that $2^n \in \Omega(n^{\log_2 \log_2 n})$, thus I want to show, that there is a value $n_0$ from which on $2^n$ will always be bigger than $n^{\log_2 \log_2 n}$.
$2^n \geq n^{\log_2 \log_2 n}$
I have thought about induction and limiting values, however it didn't seem to work. I have also thought about showing that $2^n$ is growing faster through:
$\lim_{n \rightarrow \infty} \frac{2^n}{n^{\log_2 \log_2 n}}$
And using L'Hospital or is there an approach which is more obvious?