Why is $(\log(n))^{99} = o(n^{\frac{1}{99}})$
$\qquad \begin{align}
  \lim_{x \to \infty} \frac{ (\log(x))^{99} }{x^{\frac{1}{99}}}
&amp;= \lim_{x \to \infty} \frac{ (99^2)(\log(x))^{98} }{x^{\frac{1}{99}}} \\
&amp;= \lim_{x \to \infty} \frac{ (99^3) \times 98(\log(x))^{97} }{x^{\frac{1}{99}}} \\
&amp;\vdots \\
&amp;= \lim_{x \to \infty} \frac{ (99^{99})\times 99! }{x^{\frac{1}{99}}} \\
&amp;= 0
\end{align}$

I used L'Hôpital's rule law in each conversion assuming natural logarithm.