Mathematical Biscuit I

 Can you find two irrational numbers $a,b$ such that $a^b$ is rational? Surprisingly, yes and the argument is very easy. 

If $\sqrt{2}^{\sqrt{2}}$ is rational then we are done.

If $\sqrt{2}^{\sqrt{2}}$ is irrational then $$(\sqrt{2}^{\sqrt{2}})^{\sqrt{2}}=\sqrt{2}^{\sqrt{2} \sqrt{2}}=\sqrt{2}^2=2$$ and we are done :)

PS: It turns out that $\sqrt{2}^{\sqrt{2}}$ is indeed irrational (infact transcendental). This is a consequence of an advanced result called Gelfond-Schneider Theorem. However this was irrelevant to our argument.