If I have $n^{1.161}$ and $n^{1.58}$, how do they compare in terms of time complexity relative to $n\log n$?
I only know that $n\log n$ is between $n$ and $n^2$.
I would probably factor out $n$ such that I will get $\log n$ vs $n^a$ where $0<a<1$, and that's as far I know.
As $n\to\infty$, $n\log n$ is between $n$ and $n^a$ for any $a>1$.
As you suggests, we can drop a factor of $n$ and simplify the above statement to: Given $\epsilon>0$. Then as $n\to\infty$, $\log n$ is between $1$ and $n^\epsilon $. The observation that $\log$ grows slower than any (positive) power is jut the reverse to the observation that the exponential function grows faster than any power.