Why are there infinitely many positive $k$ such that $\lfloor \frac{n^k}{k}\rfloor$ is odd? Give the postive integer $n>1$, there exist infinite positive integer $k$
such that $\lfloor \dfrac{n^k}{k}\rfloor$ is odd 
Maybe we can Use Euler's theorem,$$n^{\phi{(k)}}\equiv 1\pmod k$$
let
$$n^{\phi{(k)}}=rk+1\Longrightarrow n^k=\left(kr+1\right)^{\frac{k}{\phi{(k)}}}\Longrightarrow \dfrac{n^k}{k}=\dfrac{\left(kr+1\right)^{\frac{k}{\phi{(k)}}}}{k}$$I don't think this last part is relevant to understanding the exercise as stated, but I may be wrong. 

It seems the following.
If $n$ is odd, then we have an infinite sequence $\{n, n^n, n^{n^n}, \dots\}$ of suitable $k$’s.
