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?

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    $\begingroup$ 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) $\endgroup$ – Peter May 2 '20 at 16:29
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    $\begingroup$ @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). $\endgroup$ – Gae. S. May 2 '20 at 16:31
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    $\begingroup$ Yes, indeed it is a stronger result :) Of course, I fully expect that $r^{r^{r^r}}$ is also transcendental. $\endgroup$ – Dark Malthorp May 2 '20 at 16:33
  • $\begingroup$ @DarkMalthorp how do u prove $r^r \not \in \mathbb{Q}$? $\endgroup$ – mathworker21 May 3 '20 at 19:35
  • $\begingroup$ en.wikipedia.org/wiki/… - the techniques for the proof by infinite descent here can be generalized in a straightforward way $\endgroup$ – Dark Malthorp May 3 '20 at 19:39

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