Existence of Rational Powers of Irrational Numbers/Proof 2

Theorem
There exist irrational numbers $a$ and $b$ such that $a^b$ is rational.

Proof
Given that $2$ is rational and $\sqrt 2$ is irrational, consider the number $q = \sqrt 2^{\sqrt 2}$.

We consider the two cases.


 * $(1): \quad$ If $q$ is rational then $a = \sqrt 2$ and $b = \sqrt 2$ are the desired irrational numbers.
 * $(2): \quad$ If $q$ is irrational then $q^{\sqrt 2} = \left({\sqrt 2 ^{\sqrt 2}}\right)^{\sqrt 2} = \sqrt 2 ^{\left({\sqrt 2}\right) \left({\sqrt 2}\right)} = \sqrt 2^2 = 2$ is rational, so $a = q = \sqrt2 ^{\sqrt 2}$ and $b = \sqrt 2$ are the desired irrational numbers.