Do there exist complex algebraic $α,β$ such that $α^β=π$ or $α^β=e$? Given the algebraic operations and complex exponentiation $(a+bi)^{c+di}$ and logarithm, is it possible to derive $\pi$ and $e$?
If one is derivable then so should be the other, as $e^\pi = (-1)^{-i}$. I however don't think either are. I'd be interested to be proved wrong though.
An elaboration of the rules: no trigonometry, solution must be expressible using a finite number of terms, and $α$ and $β$ are complex, algebraic numbers.
 A: If we allow natural logarithms, then $\pi=-i\cdot\ln(-1)$. If not, and/or if we restrict the domain to algebraic reals $(\mathbb{A}\cap\mathbb{R})$, then I believe the question to be still open/undecided. We know, from the Gelfond-Schneider theorem, that $a^b$ is transcendental, for a, b $\in\mathbb{A}$, with a $\neq0,1$, and b $\not\in\mathbb{Q}$. But since algebraics are countable, whereas transcendentals aren't, it follows that almost all $\mathbb{T}$ numbers are not of the form $a^b$, with a, b $\in\mathbb{A}$, so there would be nothing surprising if it somebody were to ultimately prove that π and e could not be written in this manner. If such remarkable expressions for e or π would had been discovered by now, we'd undoubtedly have heard of them; and if a proof of their impossibility had already been established, it would've been as famous and well-known a theorem as that of Gelfond and Schneider, which I mentioned just earlier. That's why I said that the question is still open. Hope this answer helps to set things a bit more in perspective.
