Is $\sqrt{2}^{\sqrt{2}}$ rational?

Question

In §2.2 of her essay on mathematical morality, Eugenia Cheng includes the following example:

6. Why is it possible for an irrational to the power of an irrational to be rational?

   Here is a nice little proof that it is possible:

   Consider $\sqrt{2}^{\sqrt{2}}$.

   <p>If it is rational, we are done.</p>
   
   <p>If it is irrational, consider
   $$ \left(\sqrt{2}^{\sqrt{2}}\right)^\sqrt{2} = \sqrt{2}^2=2.$$</p>
   

However, as Cheng notes, this doesn't tell us whether $\sqrt{2}^{\sqrt{2}}$ itself is rational or not. So which is it?

Answer 1

It is irrational (in fact it is transcendental). This can be shown using the Gelfond-Schneider theorem

By the theorem, where $a$ and $b$ are both algebraic with $a \not \in \{0,1\}$ and $b$ irrational, $a^b$ is transcendental. Transcendental numbers are necessarily irrational.