Why is $\log_{2}n = O(n^{0.00001})$?
$n^{0.00001}$ is not approximately $1$. $n^{0.00001}$ goes to infinity as $n \to \infty$.

You can see that $\log_2 n = o( n^{0.00001} )$ by taking the limit of their ratio:
$$
\lim_{n \to \infty} \frac{\log_2 n}{n^{0.00001}} = 
\lim_{n \to \infty} \frac{n^{-1}}{0.00001 \cdot n^{0.00001} \cdot n^{-1}} =
\lim_{n \to \infty} \frac{100000}{n^{0.00001}} = 0.
$$

This tells you that you can pick any value of $c&gt;0$, for example $c=1$.
Now you just need a value $n_0$ such that $n^{0.00001} - \log_2 n \ge 0 \; \forall n \ge n_0$.

The derivative of $n^{0.00001} - \log_2 n$ is 
$n^{-1}( 0.00001 \cdot n^{0.00001} - 1)$ which is non-negative as soon as $ n^{0.00001} \ge 100000$, i.e., for $n \ge 10^{5 \cdot 10^5}$.

You can then pick any value of $n_0$ such that $(n_0)^{0.00001} - \log_2 n_0$ is non-negative and $n_0 \ge 10^{5 \cdot 10^5}$. For example $n_0 = 2^{10^{7}}$. Indeed:


$(n_0)^{0.00001} = 2^{100} = 1024^{10} &gt; 1000^{10} = 10^{30} &gt; 10^7 = \log_2 n_0$; and
$n_0 = 1024^{10^6} &gt; 1000^{10^6} = 10^{3 \cdot 10^6} &gt; 10^{5 \cdot 10^5}$.