When does $k^k = \mathrm{poly}(n)$?
More specifically, does $$ k^k = \mathrm{poly}(n) \impliedby k = \mathcal O\left( \frac{\log n}{\log \log n} \right) $$ hold? I saw that $$ k! = \mathrm{poly}(n) \impliedby k = \mathcal O \left( \frac{\log n}{\log \log n} \right) $$ holds, though struggle to prove that either.
Edit: I should have been more clear that $k$ depends on $n$. The problem came up in the study of randomised algorithms on graphs, where $k$ is path size and $n$ is the number of vertices. It came up in a discussion on cases for an algorithm to be RP (with little detail on these "trivialities" I do not get).
Let $$ k \leq C \, \frac{\log n}{\log\log n} $$ where $C \geq 2$ is an arbitrary constant. Then, for large enough $n > e^e$, \begin{align*} k^k &= e^{k\log k} \leq \exp\left( \frac{C\log n}{\log\log n} \times \log\left( \frac{C\log n}{\log \log n} \right) \right) \\ &\leq \exp\left( \frac{C\log n}{\log\log n} \times \log (C\log n) \right) \\ &\leq \exp\left( \frac{C\log n}{\log\log n} \times C \log \log n \right) \\ &= \exp \left( C^2 \log n \right) \\ &= n^{C^2} = \mathrm{poly}(n) \end{align*} as required.