I don't know whether or not $\displaystyle x = \sqrt{2}^\sqrt{2}$ is rational or not, but there is an interesting proof in which this quantity figures.
Problem: Can $\displaystyle a^b$ be rational if a and b are irrational?
Consider $\displaystyle x = \sqrt{2}^\sqrt{2}$ . If x is rational then we have our example. If not, and x is irrational, then consider $\displaystyle x^{\sqrt{2}} = \sqrt{2}^2 = 2$, and we again have an example of rational $\displaystyle a^b$ with irrational a, b. So whether x is rational or not, we have the required example.
This is an example of (to me) an absolutely infuriating proof. We know either x or $\displaystyle x^\sqrt{2}$ is an example of rational $\displaystyle a^b$ with irrational a and b, but we don't know which, and the proof gives us no clue!
This proof is discussed in "The Princeton Companion to Mathematics".