Can an irrational number raised to an irrational power be rational? Can an irrational number raised to an irrational power be rational?

If it can be rational, how can one prove it?
There is a classic example here. Consider $A=\sqrt{2}^\sqrt{2}$. Then $A$ is either rational or irrational. If it is irrational, then we have $A^\sqrt{2}=\sqrt{2}^2=2$.
Let me expand orangeskid's answer, both because I think it teaches us something useful, and it might be the easiest elementary proof for this question.
The proof that $ \sqrt{2} $ is irrational is well-known, so I will not repeat it here.
But there's a proof just as simple showing that $ \log 3 / \log 2 $ is irrational.  Suppose on contrary that $ \log 3 / \log 2 = p / q $ where p and q are integers.  Since $ 0 &lt; \log 3 / \log 2 $, we can choose $ p $ and $ q $ both as positive integers.  The equality then rearranges to $ 3^q = 2^p $.  But here, the left hand side is odd and the right hand side is even, so we get a contradiction.
That gives a positive answer to the original question:
$$
\big(\sqrt{2}\big)^{2 \log 3 / \log 2} = 3
$$
Thanks to Rand al'Thor, who mentioned this problem in SE chat and thus inspired this answer.