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

Question

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.

Answer 1

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 $\exp - 1$.