Is $e^a$ always algebraic for transcendental $a$?

Question

From Lindemann–Weierstrass theorem, it is known that $e^a$ for non-zero algebraic $a$ is always transcendental. But if $a$ is transcendental, is the opposite ($e^a \in \mathbb A$) always true?

Answer 1

No. There are uncountably many transcendental $a\in\Bbb R$, each with a different $e^a$, of which only countably many are algebraic.