How can I prove this question?

Question

We know that it is not proved that $e^e$ is transcendental, so neither is the number that $e^{e\sqrt{2}}$. My question is, if one turns out to be, how can it be proved that the other is? Because there may be some connection between the two...

Answer 1

Partial answer:

Let $a = \mathrm{e}^\mathrm{e}$.

If $a$ turns out to be algebraic (i.e., not transcendental), then $\mathrm{e}^{(\mathrm{e} \sqrt{2})} = a^\sqrt{2}$ is an algeraic number that is not zero or one, raised to an irrational algebraic power. Then, by the Gelfond-Schneider theorem, $\mathrm{e}^{(\mathrm{e} \sqrt{2})}$ is transcendental.

By a similar argument, if $a^\sqrt{2}$ turns out to be algebraic, then the Gelfond-Schneider theorem applies to $(a^\sqrt{2})^\sqrt{2} = a^2$, making $a^2$ transcendental. Then $a$ is transcendental.

So at most one of $\mathrm{e}^{\mathrm{e}}$ and $\mathrm{e}^{(\mathrm{e} \sqrt{2})}$ is algebraic. Equivalently, at least one of $\mathrm{e}^{\mathrm{e}}$ and $\mathrm{e}^{(\mathrm{e} \sqrt{2})}$ is transcendental.

The situation with (non-integer) powers of transcendental numbers is less well understood. I would be surprised if there were any known consequences for $a^\sqrt{2}$ after assuming $a$ transcendental or any known consequences for $a$ after assuming $a^\sqrt{2}$ transcendental.