Rationality of a power with irrational exponent The following fact is known:
If $a$ and $b$ are both irrational numbers, then $a^b$ can be a rational number.
Proof. Suppose $a^b$ is always irrational. Then $\sqrt{2}^\sqrt{2}$ is an irrational number, which in turn implies that  $\left(\sqrt{2}^\sqrt{2}\right)^\sqrt{2}$ must also be irrational. However:
$$
\left(\sqrt{2}^\sqrt{2}\right)^\sqrt{2} = \sqrt{2}^{\sqrt{2}\times\sqrt{2}} = 2
$$

Now assume $a$ is a rational number and $b$ is irrational. I wonder if  $a^b$ can be a rational number. Any way to prove, or a reference maybe?

You can use the Gelfond-Schneider theorem to prove that $a^b$ is not only irrational, but transcendental. The theorem states that:
If $a$ and $b$ are algebraic numbers with $a ≠ 0,1$ and $b$ non-rational, then any value of $a^b$ is a transcendental number.
This not only gives rise to trivial results such that $2^{\sqrt 2}$ is irrational, but to some more interesting ones as well. For instance, by taking $a=-1$ and $b=-i$, where $i=\sqrt {-1}$, we can obtain the following result:
$$
(-1)^{-i} = \left( e^{i \pi} \right)^{-i} = e^\pi,
$$
showing that $e^\pi$ is transcendental.
