elementary number theory - Prove that if $n geq 2$, then $sqrt[n]{n}$ is irrational. Hint, show that if $n geq 2$, then $2^{n} > n$.

Prove that if $n \geq 2$, then $\sqrt[n]{n}$ is irrational. Hint, show that if $n \geq 2$, then $2^{n} > n$.

So, my thought process was that I could show that $2^{n} > n$ using induction, but I'm not sure how that helps to solve the original problem. Showing $2^{n} > n$ means I could take the $n^{th}$ root of each side, giving me $2 > \sqrt[n]{n}$, but that's not quite showing that it's irrational.

My other thought was proof by contradiction, I could start assuming that it's rational -- ie that there exists some integers $r$ and $s$ such that $\sqrt[n]{n} = r/s$. Then I could say that $n = r^{n} / s^{n}$. But I have no way of showing that this is impossible for any $n \geq 2$.

Any hints as to what a good direction to start in would be appreciated.

Answer

Suppose there is a rational number $a/b$ such that gcd$(a,b)=1$ and $a^n/b^n=n.$ Then $a^n=b^nn.$ So $a^n$ divides $b^nn$. Since gcd$(a,b)=1 \implies$ gcd$(a^n,b^n)=1$, we know $a^n$ must divide $n$. Thus, $n \geq a^n$. But note that the integer $a$ must be at least $2$ since $(1/b)^n$ clearly cannot be $n$. Combining with the hint gives a contradiction.