TITLE: How to prove that $k^{\pi}$ is not an integer for any integer $k\geq 2$?
QUESTION [4 upvotes]: I strongly suspect that $k^{\pi}$ is not an integer for any integer
$k\geq 2$ (for otherwise this would be a famous result of which I am not aware). But how does one prove this?

The answer to this question cannot be generalised to any integer $k$, only $k=5.$ Are there better known properties of $\pi$ which  answer my above question? For example, if there were such a $k$, then we would have $\log_n k = \pi$ for some integers $k,n.$ Is there any general reason or theorem that says that this cannot be possible?

REPLY [7 votes]: I believe this is an open problem. It's amazing that number theory contains so many problems that are this simple to state yet impossible to prove so far!
It's worth remarking that the assertion that $k^\pi$ is never an integer when $k\ge2$ is an integer would follow from Schanuel's conjecture in the following way: set $z_1=\pi\log k$, $z_2=i\pi$, and $z_3=\log k$, so that $\{z_1,z_2,z_3\}$ is linearly independent over $\Bbb Q$ since $\pi$ is known to be irrational. Schanuel's conjecture then implies that the field
$$
\Bbb Q(z_1,z_2,z_3,e^{z_1},e^{z_2},e^{z_3}) = \Bbb Q(\pi\log k,i\pi,\log k,k^\pi,-1,k) \subset \Bbb Q(i,\pi,\log k,k^\pi)
$$
has transcendence degree at least $3$ over $\Bbb Q$; since $i$ is algebraic, this implies that ($\pi$ and $\log k$ and) $k^\pi$ must be transcendental and in particular not an integer. Indeed, the same logic shows that Schanuel's conjecture implies that $r^\pi$ is always transcendental (hence always irrational, hence never an integer) for any positive rational number $r\ne1$.
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