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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.

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Do we really need a tag as subjective as beautiful-proofs? – Asaf Karagila Mar 11 '14 at 13:10
@AsafKaragila I was wondering, too. But then again, mathematicians do tend to find a common sense of beauty. At least in my experience it’s not that diverse. I don’t think it’d be as problematic as say “(good-music)” would be. Maybe a meta discussion on this? – k.stm Mar 11 '14 at 13:15
Incidentally, it's known today that $\sqrt 2^{\sqrt 2}$ is irrational. See this. – David Mitra Mar 11 '14 at 13:22
@AsafKaragila Probably not. But as my question is explicitly looking for something as subjective as a beautiful proof I thought I should grab the opportunity to create my first tag... – user1729 Mar 11 '14 at 14:54

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