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

14$\begingroup$ Why are you interested in these particular numbers? $\endgroup$ – cfranc Sep 27 '10 at 14:08

8$\begingroup$ No doubt that a proof of irrationality of one of these numbers would be a monument of the human intelligence... But isn't a bit sad, such a big effort to prove something that everybody would believe true? What I would really like to see is a proof of rationality of at least one of these combinations of $\pi$ $e$ and $\gamma$. $\endgroup$ – Pietro Majer Sep 27 '10 at 17:26

10$\begingroup$ Pietro, why would it be sad to prove something people believe? It happens all the time! More often than not (but not always) longstanding conjectures which are solved turn out to be true in the way that they were conjectured. $\endgroup$ – KConrad Oct 2 '10 at 16:28

5$\begingroup$ Pietro said that it would be sad if effort were put into such things (rather than into something more enlightening or useful). I agree. $\endgroup$ – Paul Taylor May 3 '13 at 20:29

5$\begingroup$ @PaulTaylor Don't you think that the rationality of $\pi e$ would be very enlightening and useful? $\endgroup$ – Oksana Gimmel May 3 '13 at 20:50
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).

9$\begingroup$ You raise a nice question! (Though of course an answer 'yes' would be a lot nicer than 'no'!) $\endgroup$ – Stefan Kohl May 3 '13 at 20:54

10$\begingroup$ It is mentioned on the Russian Wikipedia page Open mathematical problems. A very similar question was discussed at math.stackexchange.com/questions/13050/eee79andultrafinitism $\endgroup$ – Oksana Gimmel May 3 '13 at 22:02

2$\begingroup$ ok, pi^pi^pi^pi is a hell of a lot bigger than I thought it was. Should still work though if you have a good computer and enough time. $\endgroup$ – user85798 Apr 21 '14 at 19:45

3$\begingroup$ @VladimirReshetnikov Oh, if it actually is an integer then of course this wouldn't work. I'm assuming it's not an integer. (I see no reason why we would get
x.00000000000...
) $\endgroup$ – user85798 Apr 21 '14 at 20:49 
11$\begingroup$ $\pi^{\pi^{\pi^{\pi}}}$ has over a hundred quadrillion digits. It would take more than two exabytes of storage just to write down the integer part of that number. $\endgroup$ – Daniel McLaury Jan 31 '15 at 21:07
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), 191202.
W. D. Brownawell, "The algebraic independence of certain numbers related by the exponential function," J. Number Theory 6 (1974), 2331.
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.