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?
 A: Assume $n>2$.  Prove that $n$ has to be an $n-2$ power therefore $n\ge2^{n-2}$.  Show that this is contradicted except for sufficiently small $n$.
A: As you figured out, $n = 1, 3, 4$ are solutions. Of course, $n \neq 2$ because dividng by $0$ is not defined. Now, we see $n = 5$ is not a solution.
For better understanding, notice that $n^{\frac{1}{n - 2}}$ is $(n - 2)$-th root of $n$. So, for $n = 5$ we have $\sqrt[3]5$, for $n = 10$ we have $\sqrt[8]10$, for $n = 100$ we have $\sqrt[98]100,$ and so on. It's intuitively obivous that all these roots are between $1$ and $2$ (and thus for $n \geq 5$ is $\sqrt[n - 2]n \not\in \mathbb{N}$). Let's prove this.
For $n \in \{5, 6, ...\}$ we have to prove $n^{\frac{1}{n - 2}} > 1 \iff n > 1^{n - 2} = 1$ which is obviously true. Now, we will easly prove by induction that for any $n \in \{5, 6, ...\}$ is $n < 2^{n - 2} \iff 4n < 2^n.$
Again, for $n = 5$ we have $4 \cdot 5 < 2^5 = 32$ which is true. Let's suppose $4m < 2^m$ is true for some $m \in \{6, 7, ...\}.$ But if this is true then also $2 \cdot 4m < 2 \cdot 2^m$ is true. Notice that the last equation is just case when $m := m + 1$. We proved that if equation is valid for $m$ it's also valid for $m + 1$ (and infact, if equation is false for $m$ then it's also false for $m + 1$). We proved that equation is true for $m = 5$. Then it's true for $m = 6$. Then it's true for $m = 7$, etc.
Therefore only solutions are $n = 1, 3, 4.$
A: 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.
