# If $r\in\mathbb{Q}\setminus\mathbb{Z}$ is it possible that $r^{r^{r^r}}\in \mathbb{Q}$?

It's straightforward to prove that $$r^r\notin\mathbb{Q}$$, which furthermore allows us to use the Gelfond-Schneider theorem to prove that $$r^{r^r}\notin\mathbb{Q}$$. Is it true that $$r^{r^{r^r}}\notin\mathbb{Q}$$? It seems like it ought to be true, though my guess would be that this is an open problem. If not, does anyone know a proof? And if so, do there exist any discussions of it in the literature?

• I would also guess that this is an open problem. It is even unknown whether the power-tower is an integer for $\ r=\pi\$ (which would however be a miracle) May 2, 2020 at 16:29
• @darkmalthorp For the record, Gelfond-Schneider says that $r^{r^r}\notin\overline{\Bbb Q}$, not just $\notin\Bbb Q$ (though I presume you meant that).
– user239203
May 2, 2020 at 16:31
• Yes, indeed it is a stronger result :) Of course, I fully expect that $r^{r^{r^r}}$ is also transcendental. May 2, 2020 at 16:33
• @DarkMalthorp how do u prove $r^r \not \in \mathbb{Q}$? May 3, 2020 at 19:35
• en.wikipedia.org/wiki/… - the techniques for the proof by infinite descent here can be generalized in a straightforward way May 3, 2020 at 19:39