$n^a$ integral for all integer $n$ implies $a$ integral Let $a>0$ be a real number, such that for all integers $n\geq 1$: $n^a \in \mathbb N$
Show that $a$ must be an integer.

It's not difficult to show this when $a$ is a rational number: $2^\frac{p}{q}$ is irrational when the fraction is in lowest terms and $q \neq 1$.
When $a$ is irrational, then for all $n\geq 1$, there exists $m_n\in \mathbb N^*$, such that:
$$a = \frac{\log m_n}{\log n}; \quad \text{$m_n$ is not a power of $n$}$$
I think considering $n=2,3$ is enough to show a contradiction, but I can't seem to find it. This is what I get:
$$a = \frac{\log p}{\log 2} = \frac{\log q}{\log 3}$$
$$p = 2^{\log q/\log 3} $$
I think the RHS is irrational when $q$ is not a power of $3$, but I can't prove it. The closest thing I have to a solution is this answer on a similar question. But it uses an unproven conjecture, and I was hoping for a more elementary proof.
 A: The idea is decreasing the exponent $a$ below $0$ by taking differences.
For every function $f$, let $(\Delta f)(x)=f(x+1)-f(x)$ be the usual forward difference.
$Lemma$. If $f(x)$ is a $k$ times differentiable function on $[n,n+k]$ then there is some $\xi\in(n,n+k)$ such that
$$
f^{(k)}(\xi) = (\Delta^kf)(n) = \sum_{\ell=0}^k (-1)^{k-\ell}\binom{k}{\ell}f(n+\ell).
$$
$Proof$. Induction on $k$. For $k=1$ this is exactly Lagrange's mean value theorem.
If the Lemma holds true for $k-1$ then apply it to the function $g(x)=\Delta f$. With some $\zeta\in(n,n+k-1)$, and then
$\xi\in(\zeta,\zeta+1)\subset(n,n+k)$, we have
$$
(\Delta^k f) (n) = (\Delta^{k-1} g)(n) 
= g^{(k-1)}(\zeta) = (\Delta f^{(k-1)})(\zeta)
= f^{(k)}(\xi),
$$
the lemma has been proved.
Now suppose that $a$ is not an integer; let $k$ be a positive integer with $k-1<a<k$. Applying the lemma with an arbitrary positive integer $n$,
$$
a(a-1)\dots(a-k+1) \, \xi^{a-k} = \sum_{\ell=0}^k (-1)^{k-\ell}\binom{k}{\ell} (n+\ell)^a.
$$
If $n$  is sufficiently large then the LHS is a number between $0$ and $1$. But we have some integer on the RHS, contradiction.
