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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).

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    Assuming $poly(n)$ represents the class of all polynomials in $n$ then $k^k$ is a constant "from $poly(n)$ point of view" namely : $k^k = 0\cdot n^m + 0\cdot n^{m-1} + ... + 0\cdot n + k^k$ thus $k^k \in poly(n)$. – 2017-01-19
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    We have that if $k=\log(n)$ then $k^k=n^{\log(\log(n))}$ for large $n$ for example $n=10^{100000000}$ we have that $\log(\log(n))\approx 19.25$ hence you can compare it to $n^{20}$ for example.This is just practical though,not mathematical. – 2017-01-19

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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.