Compare log^k(n) with n^(1/2)
Yes. Using L'Hospital:

$\lim_{n \rightarrow \infty} \frac{\ln^k (n)}{\sqrt n} = \lim \frac{k \ln^{k - 1} (n) \frac{1}{n}}{ \frac{1}{\sqrt n }} = \lim \frac{ k \sqrt n \ln^{k -1 } (n) }{ n } = \lim \frac{k \ln^{k-1} (n)}{\sqrt n} = \dots = \lim \frac{k \cdot (k -1) \cdots  2 \ln (n) }{ \sqrt n} = \lim \frac{k! \frac{1}{n}}{\frac{1}{\sqrt n}} = \lim \frac{k!}{\sqrt n} = 0$

I hope I got it right, quite some time since I've used it. I^ve left out the $\frac{1}{2}$ from the $\sqrt n$