If $n^s\in \mathbb{N}$ for all $n\in\mathbb{N}$, must $s\in\mathbb{N}?$

Question

My friend and I were discussing this, and while it looks obviously true, weren't very successful in getting anywhere with it. I'm not sure whether I lack the tools to solve this or whether I am missing something obvious. Any insight would be much appreciated.

Answer 1

Partial solution: if $s$ is not an integer, then it has to be transcendental.

1. It cannot be a rational number. Otherwise say $s={p\over q}$ with $q>1,\gcd(p,q)=1$, and $N=2^s\in\Bbb N$ $$q\ |\ v_2(N^q)=v_2(2^{p})=p$$ which contradicts $q>1,\gcd(p,q)=1$. Here $v_2$ stands for the $2$-adic order function.

2. Now since $s$ is not a rational, it follows from Gelfond-Schneider theorem that it cannot be an algebraic number either. Otherwise $2^s$ would be transcendental.

One can observe that so far only the fact that $2^s\in\Bbb N$ has been used. This condition alone is not sufficient to rule out transcendental numbers as was pointed out by Grant B., e.g. $2^{\log_2(3)}=3$.