I am reading a proof involving the existence of a property in a tournament (a directed complete graph).
To make the proof work, we need to have $n^ke^{-n/2^k}<1$. Here $n$ is the order of the tournament and $k$ is the order of some sub tournament.
The proof goes on to say that if $n>k^22^{k+1}$, then $n^ke^{-n/2^k}<1$, but I am not sure how to derive this bound on $n$? I have that $$\begin{align*} n^ke^{-n/2^k}&<1\\ e^{-n/2^k}&<n^{-k}\\ -\frac{n}{2^k}&<-k\ln{n}\\ \frac{n}{\ln{n}}&>k2^k \end{align*}$$ Since $\ln{x}$ is an increasing function and $n\ge 2$ we have $$ \frac{n}{\ln{2}}\ge\frac{n}{\ln{n}}>k2^k $$ which yields $$ n>k2^k\ln{2}. $$ So I am guessing that since $2>\ln{2}$ and $k>1$, we have $n>k^22^{k+1}>k2^{k+1}>k2^k\ln{2}.$
Why not state the lower version of the bound, $k2^k\ln{2}$, instead of the seemingly contrived higher bound, $k^22^{k+1}$?