How can I prove that $\log^k(n) = O(n^\epsilon)$? How can I prove that $\log^k(n) = O(n^\epsilon)$?
(for any $\epsilon > 0 $ and for any integer and positive $k$).
I tried to do it by - definition , and to prove that exists $c,n_0 > 0$ so that for any $n > n_0 $ exists: $c \cdot n^\epsilon > \log^k(n) $, but I don't have idea how to continue for this point.  
I will be happy to listen ideas :).

By definition we have
$$\log^k(n) = n^\epsilon\cdot \frac{\log^k(n)}{n^\epsilon}$$
with
$$\frac{\log^k(n)}{n^\epsilon}\to 0$$
(refer to Evaluation $\lim_{n\to \infty}\frac{{\log^k n}}{n^{\epsilon}}$)
therefore
$$\log^k(n) = o(n^\epsilon)$$
