Question

What is known about irrationality of $\pi e$, $\pi^\pi$ and $e^{\pi^2}$?

Answer 1

I believe most such questions are still very far from being resolved.

Apparently, it is not even known if $\pi^{\pi^{\pi^\pi}}$ is an integer (let alone irrational).

Answer 2

Brownawell and Waldschmidt do have results in these directions which do not rely on Schanuel's Conjecture. The references are

M. Waldschmidt, "Solution du Huitième Problème de Schneider," J. Number Theory 5 (1973), 191-202.

W. D. Brownawell, "The algebraic independence of certain numbers related by the exponential function," J. Number Theory 6 (1974), 23-31.

The two papers independently prove results along the following lines. (The following version is taken from Brownawell.) Let $\alpha$, $\beta$, and $\gamma$ be nonzero complex numbers with $\alpha$ and $\beta$ both irrational. If $e^\gamma$ and $e^{\alpha\gamma}$ are both algebraic numbers, then at least two of the numbers $$\alpha, \beta, \gamma, e^{\beta\gamma}, e^{\alpha\beta\gamma}$$ are algebraically independent over $\mathbb{Q}$.

This theorem has several interesting consequences:

• Taking $\alpha=\beta=e^{-1}, \gamma=e^2$, we see that at least one of $e^e$ and $e^{e^2}$ must be transcendental. This was conjectured by Schneider.

• Taking $\alpha=\beta=\gamma$, we see that given any nonzero complex number $\alpha$, at least one of the numbers $e^{\alpha}, e^{\alpha^2}, e^{\alpha^3}$ must be transcendental.

• Taking $\alpha = \beta = i/\pi, \gamma=\pi^2$, we see that at least one of the following holds: (i) $e^{\pi^2}$ is transcendental, or (ii) $e$ and $\pi$ are algebraically independent.

So as a partial answer to this question, at least one of $e\pi$ and $e^{\pi^2}$ is transcendental.