Algorithm notation between two functions
$\frac{f(n)}{g(n)} = \frac{n^{1.6}}{n^{1.5}} = n^{0.1}$ is unbounded. Hence, for any $c &gt; 0$ there is $n_0 \in \mathbb{N}$ such that $n^{0.1} &gt; c$ for all $n &gt; n_0$. As a result, for any $c &gt; 0$, we have $n^{1.6} = n^{0.1} \cdot n^{1.5} &gt; c n^{1.5}$ for all $n &gt; n_0$.

Notice this does not really make use of $1.5$ and $1.6$ other than $1.6 &gt; 1.5$. In fact, using the same method and the fact that $n^\varepsilon$ is unbounded for any $\varepsilon &gt; 0$, you can prove the more general bound $n^a \in \omega(n^b)$ for any $a &gt; b \ge 0$.