$a$ transcendental $\implies a^a$ is transcendental?

Question

Suppose $a\in \mathbb{C}$ is not a algebraic number. Then is $a^{a}$ also transcendental number ?

I've not idea about how to do it. I got motivation for asking this question from the fact that $e^{i\pi}$ is rational while $e,i\pi$ both are transcendental.

Answer 1

Let us take $x=\exp\left(W(\log 2)\right)$, i.e. a solution of $x^x=2$.

Step 1. $x\not\in\mathbb{Q}$.
Assuming $x=\frac{p}{q}$ with $\gcd(p,q)=1$, we have $p^p=2^q\cdot q^p$, absurd.

Step 2. $x$ is not an algebraic number.
Assuming that $x$ is in algebraic number, the Gelfond-Schneider theorem gives that $2$ is a transcendental number. It is not, so:

$\color{red}{\text{your claim does not hold.}}$