Can an irrational number raised to an irrational power be

Can an irrational number raised to an irrational power be rational?
If it can be rational, how can one prove it?

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Ella Williams
There is a classic example here. Consider $$\displaystyle{A}=\sqrt{{2}}^{{\sqrt{{2}}}}$$. Then $$\displaystyle{A}$$ is either rational or irrational. If it is irrational, then we have $$\displaystyle{A}^{{\sqrt{{2}}}}=\sqrt{{2}}^{{{2}}}={2}$$
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Juan Spiller
Consider, for example, $$\displaystyle{2}^{{\frac{{1}}{\pi}}}={x}$$ where $$\displaystyle{x}$$ should probably be irrational but $$\displaystyle{x}^{\pi}={2}$$
More generally, $$\displaystyle{2}$$ and $$\displaystyle\pi$$ can be replaced by other rational and irrational numbers, respectively.
Vasquez

For example:
$$\sqrt{2}^{2\log_23}=3$$