I suggested the following problem to my friend: prove that there exist irrational numbers $a$ and $b$ such that $a^b$ is rational. The problem seems to have been discussed [in this question](https://math.stackexchange.com/questions/104119/can-an-irrational-number-raised-to-an-irrational-power-be-rational).

Now, his inital solution was like this: let's take a rational number $r$ and an irrational number $i$. Let's assume

$$a = r^i$$
$$b = \frac{1}{i}$$

So we have

$$a^b = (r^i)^\frac{1}{i} = r$$

which is rational per initial supposition. $b$ is obviously irrational if $i$ is. My friend says that it is also obvious that if $r$ is rational and $i$ is irrational, then $r^i$ is irrational. I quickly objected saying that $r = 1$ is an easy counterexample. To which my friend said, OK, for any positive rational number $r$, other than 1 and for any irrational number $i$ $r^i$ is irrational. Is this true? If so, is it easily proved? If not, can someone come up with a counterexample?

Let's stick to real numbers only (i.e. let's forget about complex numbers for now).