Does n^(1-1/d) always dominate log^d(n)
L'Hôpital's rule shows that for $\epsilon &gt; 0$,
$$
\lim_{n\to\infty} \frac{\log n}{n^\epsilon} =
\lim_{n\to\infty} \frac{\frac{1}{n}}{\epsilon n^{\epsilon-1}} =
\lim_{n\to\infty} \frac{1}{\epsilon n^\epsilon} = 0.
$$
In other words, $\log n = o(n^\epsilon)$ for all $\epsilon &gt; 0$.
This immediately implies that $\log^C n = o(n^\delta)$ for all $C,\delta &gt; 0$, taking $\epsilon = \delta/C$.