This question is a classic and is on Stack Exchange several times, but I am looking for some atypical answers. The basic question, as you all already know is, "Find two irrational numbers $a$ and $b$ such that $a^b$ is rational."
There are two very common answers. The first being the classic $(\sqrt{2}^{\sqrt{2}})^\sqrt{2} = 2$ argument (in which the irrationality of $\sqrt{2}^{\sqrt{2}}$ happens to be irrelevant) and the second being the $\sqrt{2}^{2\log_2(3)} = 2$ example. These are both trivial and traditional proofs, but are there any other examples not usually given? A bunch more examples would be nice. It would also be helpful to show that the two numbers $a$ and $b$ are irrational, as some of these proofs, like $\pi$ and $e$ are not elementary.
Cheers.
It is easy to establish that $x^x$ is irrational when $x$ is a positive non-integer rational number (see for instance, https://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1031393807). Consider the equation $$x^{x+n}=q\,,$$ where $n\ge 0$ is an integer and $q>0$ is a non-integer rational number (Note: A solution to this exists by the Intermediate Value Theorem).
Proof: Suppose $x$ is a rational number, then we have $$x^x=x^{-n}q\,.$$ Thus, $x^x$ is a rational number and thus $x$ must be a positive integer; but $x^{-n}q$ cannot be an integer, which is absurd.