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

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