Finding natural numbers $n$ such that $n^\frac{1}{n-2}\ $ is also a natural number 
Find all natural numbers $n$ such that $n^\frac{1}{n-2}\ $ is also a natural number.

I know that the numbers will be $1, 3$ and $4,$ but I am not quite sure how to approach the proof. Any suggestions on how can I prove this?

Just to give a different approach, let's write $n=m+2$ and look for cases where $m+2=k^m$ for some (positive) integer $k$. We see that
$$m+2=k^m\implies m+1=k^m-1=(k-1)(k^{m-1}+\cdots+1)\ge(k-1)m\implies 1\ge(k-2)m$$
This limits $m$ to be less than or equal to $1$ unless $k=1$ or $2$, so, on checking cases, we get $(m,k)=(1,3)$, $(-1,1)$ and $(2,2)$ as the only solutions, corresponding to $n=3$, $1$, and $4$, respectively.
