Show that $(\log(n))^\alpha = O(n^\beta)$
You can use the following identities:
$$
(\ln n)^a \le n^\beta \iff \ln (\ln n)^a \le \ln n^\beta \iff
a \ln \ln n \le \beta \ln n.
$$
This is trivial if $\alpha \le 0$, so we consider the case $\alpha &gt; 0$.
For simplicity substitute $t= \ln \ln n$ and $\gamma = \frac{\alpha}{\beta}$. Notice that $\gamma \ge 0$. We obtain:
$ 
\gamma t \le  e^t \iff \gamma t - e^t \le 0  
$
To show that $\alpha t \le \beta e^t$ for sufficiently large $t$, we take the derivative of $h(t) = \gamma t - e^t$:
$$
h'(t) =\gamma - e^t.
$$
This shows that for $t \ge \ln \gamma$, $h(t)$ is monotonically non-increasing.
Moreover, for $t=\ln \gamma$ we already have $h(t) \le 0$, indeed $h(\ln \gamma) =\ln \gamma-\gamma \le 0$ since the logarithm is a monotonically increasing function.
Hence, in the definition of big-Oh with $n_0$ and $c$, we can pick $n_0 = \ln \gamma = \ln \frac{\alpha}{\beta}$ and $c=1$.