How to show $e^{e^{e^{79}}}$ is not an integer

Question

In this question, I needed to assume in my answer that $e^{e^{e^{79}}}$ is not an integer. Is there some standard result in number theory that applies to situations like this?

Much later addendum: when I asked this, I didn't expect it to be an open question. If it is open, if someone could give an answer to that effect, I would accept it simply to close off the question. If some author has stated something similar as an open question, that would be helpful for me to know; I'm not a number theorist and my knowledge of the field is not very deep.

Answer 1

if $e^{e^{e^{79}}}$ is an integer then $e^{e^{e^{e^{79}}}}$ is not an integer (otherwise $e$ would be algebraic). Perhaps your arguments make sense with this number too.

Answer 2

The paper Chuangxun Cheng, Brian Dietel, Mathilde Herblot, Jingjing Huang, Holly Krieger, Diego Marques, Jonathan Mason, Martin Mereb, and S. Robert Wilson, Some consequences of Schanuel’s conjecture, Journal of Number Theory 129 (2009) 1464–1467, shows that $e,e^e,e^{e^e},\dots$ is an algebraically independent set, on the assumption of Schanuel's Conjecture. Maybe a close reading of that paper will suggest a way of applying the result to the 79-question.

Answer 3

Perhaps you could do something with the taylor series? Would be pretty nasty but youd only have to worry about the first few terms.