This is a classic (very old).
If you've seen it, please don't post the answer.
An irrational number raised to an irrational power is always irrational.
Prove it or provide a counterexample.
The "classic" solution goes like this:
Theorem: .$\displaystyle (\text{irrational})^{\text{irrational}}$ can be rational.
Proof
$\displaystyle \text{Consider }a \:=\:\sqrt{2}^{\sqrt{2}}$
There are only two possibilities:
. . (1) $\displaystyle a$ is rational.
. . (2) $\displaystyle a$ is irrational.
If (1) $\displaystyle a$ is rational, the theroem is verified.
If (2) $\displaystyle a$ is irrational, consider: .$\displaystyle a^{\sqrt{2}}$
. . an irrational number raised to an irrational power.
Then we have: .$\displaystyle a^{\sqrt{2}} \:=\:\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}} \:=\:\left(\sqrt{2}\right)^2\:=\:2$ . . . a rational number.
Either way, an irrational raised to an irrational power can be rational.
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I've always found this proof amusing.
Note that we still don't know if$\displaystyle \sqrt{2}^{\sqrt{2}}$ is rational or irrational
. . but it doesn't matter . . .