Consider the following question -Are there two irrational numbers $a$ and $b$ such that $a^{b}$ is rational?- Well, suppose $\sqrt{2}^{\sqrt{2}}$ is rational, then we are done. If we suppose it is irrational, then $\left(\sqrt{2}^{\sqrt{2}}\right)^\sqrt{2}$ is rational, and we are done again. In either case there is a solution, so the answer to the question is affirmative.
However, which one is it? One has to engage in some serious number theory in order to decide. That is, to show that $\sqrt{2}^{\sqrt{2}}$ is irrational. In fact, it is transcendental; as the number $\sqrt{n}^{\sqrt{n}}$ is for any positive integers $n$ that is not a square.
What is the simplest (elementary) proof of this fact? And the most elegant?