The following theorem has a very beautiful proof.
Theorem: There exist two irrational numbers $x$ and $y$ such that $x^y$ is rational.
Proof: If $\sqrt{2}^{\sqrt{2}}$ is rational then we are done. Otherwise, take $x=\sqrt{2}^{\sqrt{2}}$ and $y=\sqrt{2}$. Here, $x^y=\sqrt{2}^{\sqrt{2}\sqrt{2}}=2$.
In this proof the number $2$ plays an important role. This proof does not work if we replace $2$ with, for example, the number $3$. However, it does work if we replace $2$ with an arbitrary even number.
My question is therefore the following.
Does there exist a similarly beautiful proof that for every odd number $n$ there exist two towers-of-roots $x:=\sqrt{n}^{\sqrt{n}^{\sqrt{n}\cdots}}$ and $y:=\sqrt{n}^{\sqrt{n}^{\sqrt{n}\cdots}}$ such that $x$ and $y$ are both irrational but $x^y$ is rational? (The roots need not all be square roots, but they should all be the same.)
The "all be the same" stipulation is because you can go "If $(n^{1/n})^{(n^{1/n})^{n-1\pmod n}}$ is not rational then take $x=(n^{1/n})^{(n^{1/n})^{n-1\pmod n}}$ and $y=n^{1/n}$. Then $x^y=n$." Which is kinda pretty, if you ignore the notation and focus on what is going on. So I already have an answer if I relax this condition.
