Well the proof I would use is let a = e and b = i(pi).

e^(i theta) = cos theta + i sin theta (Euler's identity) thus e^(i pi) = cos pi + i sin pi = -1 + i(0) = -1

We know that e and i pi are irrational (in fact i pi isn't even a real) and -1 is rational.

Therefore there exist two numbers a and b such that both a and b are irrational but a^b is rational.

In fact log of just about anything is irrational so e^(log x) works as well for just about all rational x, but Euler's identity is cool so I wanted to use that.