We know that it possible to have $\text{irrational}^\text{irrational}=\text{rational}$.
To verify the possibility of this, we consider the expression $\sqrt{2}^\sqrt{2}$, if it is rational, then this step completes the verification. If $\sqrt{2}^\sqrt{2}$ is irrational, then we take a further step which is raising it to the power $\sqrt{2}$ which is irrational,
we get $\sqrt{2}^{\sqrt{2}^\sqrt{2}}=\sqrt{2}^2=2$ which is rational. This completes the verification.
The question that pops up into my mind is:
Is it possible to have $\text{transcendental}^\text{transcendental} =\text{algebraic}$?
I could not reach the answer with the same logic (as dealing with irrationals).
Any help would be appropriated. THANKS.
Yes: $e$ and $\ln2$ are trascendental and $e^{\ln 2}=2$. Also $e^{i\pi}=-1$.