Prove that $ln(n)^r \in o(n^p)$ for $p>0$ and $r\in \mathbb{R}$
You just need to show that
$$
\lim_{n \to \infty} \frac{\log^r n}{n^p} = 0.
$$
This is trivial if $r \le 0$ since $\frac{\log^r n}{n^p} = \frac{1}{n^p \log^{-r} n}$ and $\lim_{n \to \infty} n^p \log^{-r} n = +\infty$.
For $r&gt;0$, you can use  l'Hôpital's rule $\lceil r \rceil$ times to obtain:
$$
\lim_{n \to \infty} \frac{\log^r n}{n^p} 
= \lim_{n \to \infty} \frac{r \log^{r-1} n}{pn^{p}} 
= \dots
= \lim_{n \to \infty} \frac{\prod_{i=0}^{\lceil r \rceil-1} (r-i) \cdot \log^{r-\lceil r \rceil} n}{p^{\lceil r \rceil} n^{p}} \\
=
\frac{\prod_{i=0}^{\lceil r \rceil-1} (r-i)}{p^{\lceil r \rceil}}\cdot \lim_{n \to \infty} \frac{ \log^{r-\lceil r \rceil} n}{ n^{p}} =0,
$$
where $\frac{\prod_{i=0}^{\lceil r \rceil-1} (r-i)}{p^{\lceil r \rceil}}$ is a positive constant and
$
\lim_{n \to \infty} \frac{ \log^{r-\lceil r \rceil} n}{ n^{p}}
$
is equal to $0$ since it falls into the previous case.