# Proving $\log^n(n)=\omega(n^{\log(n)})$

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 :)

Let $$g(n)=\frac {(\log n)^n}{n^{\log n}}.$$
Then $$\log g(n)=n(\log n)-(\log n)(\log n)=(n-\log n)(\log n).$$
As long as your $$\log$$ is to a base $$B>1$$, we have $$n-\log n\to \infty$$ and $$\log n \to \infty$$ so $$\log g(n) \to \infty$$ so $$g(n)\to \infty.$$
Hint: Using $$a^b = 2^{b \log_2 {a}}$$, we see that $$(\log_2 n)^n = 2^{n\log_2 n}$$ and $$n^{\log_2 n} = 2^{(\log_2 n)^2}$$. Hence the ratio you are interested in is equal to $$2^{n\log_2 n - (\log_2 n)^2} = 2^{(\log_2{n}) (n - \log_2 n)}.$$ Can you explain why this tends to $$\infty$$ as $$n\to \infty$$?