# Is there any known transcendental $b$ such that $b^b$ is also transcendental?

Numbers such as $$e$$ and $$π$$ are known to be transcendental, however, $$e^e$$ or $$π^π$$ are not even known to be irrational, let alone transcendental.

There are infinitely many transcendental numbers $$a$$ such that $$a^a$$ is rational, namely the solution of every $$x^x = p$$ where $$p$$ is prime.

My question is: do we know of any transcendental number $$b$$ such that $$b^b$$ is transcendental?

Let $$b = 2/W(2)$$, where $$W$$ is the Lambert W function. Note that $$z = W(2)$$ satisfies $$z e^z = 2$$. If $$z$$ were algebraic, then $$e^z$$ would also be algebraic, but this would contradict Lindemann's theorem. Therefore $$z$$ is trancendental, and so is $$b$$. Now $$W(2) + \log(W(2)) = \log(2)$$, so $$\log(b) = \log(2) - \log(W(2)) = W(2)$$, and $$b^b = \exp(b \log(b)) = \exp(2)$$ which is transcendental.