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Is there a proof that the power tower of length $4$ of $e$ is irrational? Is it known whether or not $$e^{e^{e^e}}$$ is transcendental?

Rustyn
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    Generally, these things are really hard to prove. I wouldn't be surprised if there wasn't even a proof that $e^{e^e}$ was transcendental. – Joe Z. Apr 25 '14 at 19:37
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    No, and no. . . – Lucian Apr 25 '14 at 19:40
  • @JoeZ. $e^{e^{e^e}}$ is small enough that I think we can show that it's not an integer, I guess I'm wondering if someone's figured out whether or not it's a quotient of two integers. – Rustyn Apr 25 '14 at 19:41
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    @Rustyn Size doesn't matter (inthis context). Even to show that the small number $\pi$ is not the quotient of two inegers is much more involved than the simple verification that it is not an integer. – Hagen von Eitzen Apr 25 '14 at 20:16
  • @HagenvonEitzen Ok, I realize that, I was just wondering if it has been proven if its irrational, (or not). Sounds like Lucian knows that answer is "negative" – Rustyn Apr 26 '14 at 23:29
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    @JoeZ. Here is a somewhat partial result for that, based on the works of Brownawell. – Balarka Sen Jul 10 '14 at 18:22

2 Answers2

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As noted in a previous answer concerning the irrationality of some power towers, currently we have no tools to assess the irrationality of numbers whose exponent is transcendental. In particular, it could even be that $e^e$ is rational.

There exist however other similar tower constructions of irrational numbers whose rationality is known, for instance the infinite tower

$$ \sqrt 2^{{{\sqrt 2}^{{\sqrt 2}^{\ldots}}}} = \sqrt[\leftroot{-2}\uproot{2}4]{4}^{{\sqrt[\leftroot{-2}\uproot{2}4]{4}}^\ldots} = 2. $$

In particular, we have the following lemma.

Lemma. Let $n$ be a positive integer different than $1$, $2$ and $4$. Then the infinite power tower $$\sqrt[\leftroot{-2}\uproot{2}n]{n}^{{\sqrt[\leftroot{-2}\uproot{2}n]{n}}^\ldots}$$ is transcendental.

For instance, $$\sqrt[\leftroot{-2}\uproot{2}3]{3}^{{\sqrt[\leftroot{-2}\uproot{2}3]{3}}^\ldots}$$ is a transcendental number.

Klangen
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Let us denote $e_0 = 1$, $e_n = e^{e_{n-1}}$ (tower of length $n$). Now, I don't know whether any numbers beyond $e_1$ are known to be irrational or transcendental, although I would expect that none of those statements are proved. However, if one would assume the Schanuel's conjecture, then one can prove that all of those numbers are transcendental. Even more, any subset of them is algebraically independent, so they are indeed all entirely different.

Schanuel's conjecture states that if complex numbers $z_1,\dots,z_n$ are linearly independent over $\mathbb Q$, then the field $\mathbb Q(z_1,\dots,z_n,e^{z_1}, \dots, e^{z_n})$ has transcendence degree at least $n$ over $\mathbb Q$. Assuming it is known, the statement above can be proved by a simple induction: if $n=1$ then we know that $K_1 = \mathbb Q(1, e^1) = \mathbb Q(e)$ has $\mathrm{tr}\deg K_1 \geqslant 1$, so $e$ is transcendental (this is already known unconditionally). If $n>1$, then $$K_n = \mathbb Q(1, \dots, e_{n-1}, e^1,\dots, e^{e_{n-1}}) = \mathbb Q(e, \dots, e_n)$$ has $\mathrm{tr}\deg K_n \geqslant n$, which by induction means that all of $e,\dots, e_n$ are algebraically independent, Q.E.D.

Anton Fetisov
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