I am not understanding the following "equation":
$$\left(N\left(1-O\left(\frac{kt}{n}\right)\right)\right)^{m-(k-1)t}=N^{m-(k-1)t}\left(1-O\left(\frac{mkt}{n}\right)\right)$$
$k$ and $t$ are constant positive integers, $N= {{n}\choose{2}}$. For $m$ it is known:
- $m = cn^{\frac{k-2}{k-1}}, c>0$
- $\frac{m^2}{{n-kt}\choose{k}} \to 0$ for $n \to \infty$
specifically, I am not understanding how the exponent is getting absorbed as a factor into the $O$.
For large $n$, $\frac{kt}n$ is small and therefore $\left(1-A\frac{kt}n\right)^c$ is $1-cA\frac{kt}n$ plus lower-order terms provided $c$ is small compared to $n$, which in this case it is.