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?

After several years, it appears this is an open problem. As a non-number theorist, I had assumed there would be known results that would answer the question. I was aware of the difficulty in proving various constants to be transcendental -- such as $e + \pi$, which is not known to be transcendental at present.

However, I was looking at a question that seems simpler, naively: whether a number is an integer, rather than whether it is transcendental. It seems that what appeared to be possibly simpler is actually not, with current techniques.

The main motivation for asking about this particular number is that it is very large. It is certainly possible to find a pair of very large numbers, at least one of which is transcendental. But the current lack of knowledge about this particular number is even an integer shows just how much progress remains to be made, in my opinion. Any answers that describe techniques that would suffice to solve the problem (perhaps with other, unproven assumptions) would be very welcome.

  • $\begingroup$ You mean $\log(-1)=i \pi$. $\endgroup$ – Dog_69 Jun 15 '18 at 22:20
  • $\begingroup$ For complex numbers $e^{m i}$ is defined/derived to be $\cos m + i \sin m$ and therefore $e^{\pi i} = \cos \pi + i \sin \pi = -1$ so $\ln (-1) = \pi i$. So the real question is why is $e^{m i}$ defined/derived to be $\cos m + i\sin m$. And that is because that is the only way we can define $e^z$ so that $e^{z+ w} = e^ze^w $ and that $\frac {d(e^z)}{dz}= e^z$ $\endgroup$ – fleablood Jun 15 '18 at 22:31

$e^{i\pi}=-1$ by Euler’s equations, and so you can interpret it as $\log(-1)=\pi i$.

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  • $\begingroup$ You mean $\log(-1)=i\pi$. $\endgroup$ – Dog_69 Jun 15 '18 at 22:21
  • $\begingroup$ Sure sorry you have reason $\endgroup$ – Federico Fallucca Jun 15 '18 at 22:24
  • $\begingroup$ Not apologies at all :) $\endgroup$ – Dog_69 Jun 15 '18 at 22:35

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