Assume $e$ is rational. Then, there exist coprime integers $m$ and $n$, and we can choose $n$ to be positive, such that: $\displaystyle \frac m n = e = \sum_{i \mathop = 0}^\infty \frac 1 {i!}$ from the definition of [[Definition:Euler's Number/Limit of Series|Euler's number]].
Multiplying the previous infinite series for $e$ by $n!$ $\color{red}{(1) \text{Why do this — eliminate the denominators? Please edit this if wrong}}$:
$ \begin{align} n! \frac m n & = n! \sum_{i \mathop = 0}^\infty \frac 1 {i!} \\ & = \left({\frac{n!}{0!} + \frac{n!}{1!} + \frac{n!}{2!} + \cdots + \frac {n!}{n!}}\right) + \color{blue}{\left({\frac{n!}{\left({n + 1}\right)!} + \frac{n!}{\left({n + 2}\right)!} + \frac{n!}{\left({n + 3}\right)!} + \cdots}\right)} \end{align}$
$\color{green}{m\left({n - 1}\right)! - \left(\frac{n!}{0!} + \frac{n!}{1!} + \frac{n!}{2!} + \cdots + \frac{n!}{n!}\right)} $
$\begin{align} =\color{blue}{\frac 1 {\left({n + 1}\right)} + \frac 1 {\left({n + 1}\right) \left({n + 2}\right)} + \frac 1 {\left({n + 1}\right) \left({n + 2}\right) \left({n + 3}\right)} + \cdots} (*) \\
< \frac 1 {\left({n + 1}\right)} + \frac 1 {\left({n + 1}\right) \left({n + 1}\right)} + \frac 1 {\left({n + 1}\right) \left({n + 1}\right) \left({n + 1}\right)} + \cdots \\
=\sum_{i \mathop = 0}^\infty \left ({\frac 1 {n+1} }\right)^{\left({i + 1}\right)}
=\frac{\frac 1 {n+1} } {1 - \frac 1 {n+1} } = \frac 1 n \\ & \color{magenta}{< 1} \end{align}$
from [[Sum of Infinite Geometric Progression]]
Observe that $\color{green}{green}$ must be an integer, as it is composed entirely of sums and differences of integral components. Because $0 <$ RHS of $(*)$ and of the paragraph overhead, thence $0 < \color{green}{green} \color{magenta}{<1} $, a contradiction because no integers exist in $(0, 1)$.
(2) What ratifies the right to choose $n > 0$? Rational numbers can $< 0$ so $m$,$n$ can $< 0$?
(3) Purport that we aren't working with $2.7 < e < 2.8$. Instead we are working with another variable and don't know its sign — after letting $x = m/n$, then what are the the signs of $m, n$?
(4) The key comes off as isolating the blue sum and moving $\left({\frac{n!}{0!} + \cdots + \frac {n!}{n!}}\right)$ to the left
— later proving $\color{green}{green} < 1$? How can you prefigure this uncanny step?
ANY rational number can be written as $\frac{m}{n}$ with $n>0$. For example, $\dfrac{5}{-2}$ can also be written as $\dfrac{-5}{2}$. Just flip the sign.
We don't need to know the sign of $m$. We do in fact need to know $n>0$ since we are using $n!$ which is only defined for nonnegative integers.
One thing to notice is that the series converge (to $n!e$) so it's always possible to cut off after a certain point so that the remaining terms sum up to less than 1. The point here is that you do not want that point to be too far, because you want the partial sum to be still an integer. Luckily, it works because the series drop off very fast; the same technique would work for any similarly fast dropping series.