Is $e^{{e^{\ \dots\ }}^n}$ ever an integer?

Let $$n$$ be a positive integer. It is clear that $$e^n$$ is not integer because $$e$$ is transcendental (not algebraic).

Now for each positive integer $$k$$ let $$F^k(n)$$ denote the $$k$$-fold composition of $$F(n)=e^n$$.

Is $$F^k(n)$$ ever an integer?

I am also (primarily) interested in this question for compositions of the function $$F(n)=e^n-1$$.

It seems to me that the answer should be no in each case. I'd like to see a proof if it's fairly simple, or just a reference if this is known but complicated.

• I am not sure what is a good way to write the title in MathJax. Wikipedia article about tetration uses something like: $e^{e^{\cdot^{\cdot^{n}}}}$ which gives $e^{e^{\cdot^{\cdot^{n}}}}$. I have asked in the MathJax chatroom for advice. Feel free to revert to the original title, if you prefer that one. Commented Sep 28, 2022 at 5:05
• This function is widely denoted $\exp$. Writing $\exp^{\circ k}(n)$ makes the notation essentially self-defined.
– YCor
Commented Sep 28, 2022 at 6:48
• Somewhat related question on MSE, also with an answer conditional on Schanuel's conjecture. Commented Sep 28, 2022 at 11:25
• Why do you think it could be an integer and not a rational number? Commented Sep 28, 2022 at 17:54
• Commented Sep 29, 2022 at 0:10

The impossibility of this would follow from Schanuel's conjecture but I would be surprised if it was known unconditionally. Let $$q$$ be rational and let $$e_k = \exp^k(q)$$, so that $$e_0 = q$$. We will show the stronger result that all the $$e_k, k \ge 1$$ are algebraically independent over $$\mathbb{Q}$$, by induction (so in particular they are all transcendental). The base case is the unconditional result that $$e_1$$ is transcendental. In general, if we know that $$\{ e_1, \dots e_k \}$$ are algebraically independent, then $$\{ e_0, \dots e_k \}$$ are linearly independent over $$\mathbb{Q}$$, so by Schanuel's conjecture it follows that

$$\mathbb{Q}(e_0, \dots e_k, \exp(e_0), \dots \exp(e_k)) = \mathbb{Q}(e_0, \dots e_{k+1})$$

has transcendence degree at least $$k+1$$ over $$\mathbb{Q}$$. Since $$e_0 = q$$ is rational it follows that $$\{ e_1, \dots e_{k+1} \}$$ are algebraically independent, as desired.

Edit: Also, since we can replace $$\exp$$ with $$\exp - 1$$ in the statement of Schanuel's conjecture and generate the same field either way, the same is true for the iterates of $$\exp - 1$$.

• specifically I think it is open whether $e^{e^{e^e}}}$ is an integer. Commented Sep 28, 2022 at 12:42
• $e^{e^{e^{e}}}$ has only 1656521 digits, and $e^{e^{e^{e}}} = x + 0.2212029...$ where $x$ is an integer. You might be thinking of the open problem about whether $\pi^{\pi^{\pi^{\pi}}} \in \mathbb{Z}$. (The number $\pi^{\pi^{\pi^{\pi}}}$ has about $6 \cdot 10^{17}$ digits.) Commented Sep 28, 2022 at 17:03
• ... perhaps take $\pi/2$ instead of $\pi$ ... At least this is definitively no integer ;-) Commented Sep 28, 2022 at 17:51
• I'm amazed that e^e^e^e is so much smaller than pi^pi^pi^pi. Commented Sep 28, 2022 at 18:00
• There is a Stand-up Maths video about $\pi^{\pi^{\pi^\pi}}$. Commented Sep 28, 2022 at 19:45