Finding a lower bound for the expression $\ln^k(n)$
The notation $\log^k n$ most likely stands for $(\ln n)^k$ (this is a common abuse of notation).
In this case the claim is trivially when $\varepsilon=0$ true since $n^\varepsilon=1$ and $\ln^k n = \Omega(1)$.
For constant $\varepsilon&gt;0$ and $k \ge 1$ you have:
$$
\lim_{n \to +\infty} \frac{\ln^k n}{n^\varepsilon} =
\lim_{n \to +\infty} \left( \frac{\ln n}{n^\frac{\varepsilon}{k}} \right)^k = 
\left( \lim_{n \to +\infty}  \frac{\ln n}{n^\frac{\varepsilon}{k}} \right)^k =
\left( \lim_{n \to +\infty} \frac{1}{n \cdot \frac{\varepsilon}{k} \cdot n^{\frac{\varepsilon}{k}-1}}\right)^k
= \left( \lim_{n \to +\infty} \frac{k}{\varepsilon n^\frac{\varepsilon}{k}} \right)^k
 = 0,
$$
showing that $\ln^k n = o(n^\varepsilon)$, and hence that the claim is false.