The number $e$ is irrational

Assume $e\in \mathbb{Q}$. By [L3], write $e=p\mathrm{/}q$ in $\mathbb{R}$ with $p\in \mathbb{Z}$ and $q\in \mathbb{N}$, $q\ge 1$, using the canonical embeddings. Choose a natural $n\ge \max \{q,2\}$ (Every complete ordered field is Archimedean).

Put $A:=\iota (n!)(e-{\sum }_{k=0}^{n}1\mathrm{/}\iota (k!))$. Every tail term is positive, so $A>0$. Applying [L2] with $x=1$ and $N=n$, then using the factorial recurrence, gives $A\le \frac{2\iota (n!)}{\iota ((n+1)!)}=\frac{2}{\iota (n+1)}\le \frac{2}{3}<1$ because $n\ge 2$.

The number $A$ from step 2.1 is an embedded integer. Indeed, for each $0\le k\le n$, [L1] gives a natural $s_k$ with $n!=k!s_k$. Also $q!=m!q$ for the natural $m$ with $q=m+1$, and [L1] at $k=q$ gives $q!\mid n!$; hence $n!=qr$ for some natural $r$. By [L3] and multiplicativity of the embeddings, $\iota (n!)e=\widehat{pr},\frac{\iota (n!)}{\iota (k!)}=\iota (s_k),$ where $\widehat{pr}$ is the real image of the integer $pr$. Therefore $A$ is a difference of embedded integers and is itself an embedded integer.

Since the embedding preserves order, no embedded integer lies strictly between $0$ and $1$, contradicting steps 3.1 and 2.1. Therefore $e\mathrm{\notin }\mathbb{Q}$.