Proving $\log^n(n)=\omega(n^{\log(n)}) $
Hey everyone. I am trying to prove that $\displaystyle\lim_{n\to\infty}\frac{(\log n)^n}{n^{\log n}}=\infty $
(here $\log n= \log_2 n$)
I've tried proving this using squeeze theorem, since using L'Hôpital's rule in this case is quite out of question. I thought of proving this using the definition: $\log^n(n)=\omega(n^{\log(n)}) \iff \forall C>0 \ \ \exists N>0$ s.t. $ \forall n\ge N , (\log n)^n \ge C \cdot n^{\log n} $
Let $C$ be some arbitrary positive value. Then we can say that for all $n \ge 2^C \implies (\log n)^n>C^{2^C} $
and $ n^{\log n} \ge 2^{C^2} $
If $C>2$ then $C^{2^C} \ge 2^{C^2} $ (How can I elegantly prove this?) $\implies (\log n)^n \ge n^{\log(n)} \ , \ \forall n\ge 2^C$
Else, $0\lt C \lt 2 $... I'm quite lost and not sure if this direction is alright. I would love to hear your thoughts. Thank you :)