Order of $f(n) = \sqrt[\log n]{n} \cdot n^{\sqrt[\log n]{n}}$

Question

This is as far as i've gone: $$f(n) = \sqrt[\log n]{n} \cdot n^{\sqrt[\log n]{n}} \iff f(n)^{1 / \sqrt[\log n]{n}} = (\sqrt[\log n]{n})^{1 / \sqrt[\log n]{n} } \cdot n \iff f(n) = n^{\sqrt[\log n]{n}} $$

since $ \lim \limits_{x\to \infty} \sqrt[x]{x} = 1$

Now i haven't been able to proceed from there. If i divide $f(n)$ by $\sqrt[\log n]{n}$ i will prove that it's $\mathcal{O}\left(\sqrt[\log n]{n}\right)$ but i think i need to do better than that.. Help?

Answer 1

$$ \sqrt[\log n]{n} = n^{1/\log(n)} = \left(e^{\log(n)}\right)^{1/\log(n)} = e $$ So overall you have $f(n) = e \cdot n^e$

Answer 2

Observe that $$ \log\left(\sqrt[\log n]{n}\right) = \log\left(n^{1 / \log n}\right) = 1 $$

Answer 3

One may write, as $n \to \infty$, $$ f(n) = \sqrt[\log n]{n} \cdot n^{\sqrt[\log n]{n}}=n^{\large \frac1{\log n}}\cdot n^{\large n^{\frac1{\log n}}}=e^{\large \frac{\log n}{\log n}}n^{\large e^{\frac{\log n}{\log n}}}=e \cdot n^{\large e^1}\sim e \cdot n^{e}. $$