[proofplan]
We first rule out the possibility that $e$ is an integer by proving $2 < e < 3$. Assuming $e = p/q$ is rational with denominator $q \ge 2$, we multiply the defining [series](/pages/Series) by $q!$ and split it into a finite integer part and a positive tail. The tail is bounded above by a geometric series and lies strictly between $0$ and $1$, contradicting the integrality of $q!e$.
[/proofplan]
[step:Show that $e$ lies strictly between $2$ and $3$]
From the defining series,
\begin{align*}
e
= \sum_{j=0}^{\infty} \frac{1}{j!}
= 1 + 1 + \sum_{j=2}^{\infty} \frac{1}{j!}
> 2.
\end{align*}
For each integer $j \ge 2$, the product $j! = 1 \cdot 2 \cdots j$ contains the factor $2$ and then $j-2$ additional factors each at least $2$, so $j! \ge 2^{j-1}$. Hence
\begin{align*}
\sum_{j=2}^{\infty} \frac{1}{j!}
\le \sum_{j=2}^{\infty} \frac{1}{2^{j-1}}
= \sum_{m=1}^{\infty} \frac{1}{2^m}
= 1,
\end{align*}
where $m = j-1$ is the reindexed summation variable and the final equality is the geometric series formula. Since the term with $j = 3$ satisfies $1/3! < 1/2^{2}$, the inequality for the tail is strict. Therefore
\begin{align*}
2 < e < 3.
\end{align*}
[/step]
[step:Assume a rational representation with denominator at least $2$]
Suppose, for contradiction, that $e$ is rational. Choose $p, q \in \mathbb{N}$ with $e = p/q$. Since $2 < e < 3$, the number $e$ is not an integer, so every rational representation of $e$ in lowest terms has denominator $q \ge 2$. For such a representation,
\begin{align*}
q!e = q! \frac{p}{q} = (q-1)!p \in \mathbb{N}.
\end{align*}
[/step]
[step:Split $q!e$ into an integer part and a positive tail]
Define
\begin{align*}
N &:= \sum_{j=0}^{q} \frac{q!}{j!}, \\
x &:= \sum_{j=1}^{\infty} \frac{q!}{(q+j)!}.
\end{align*}
Then
\begin{align*}
q!e
= q! \sum_{j=0}^{\infty} \frac{1}{j!}
= \sum_{j=0}^{q} \frac{q!}{j!} + \sum_{j=q+1}^{\infty} \frac{q!}{j!}
= N + x.
\end{align*}
For each integer $j$ with $0 \le j \le q$, the quotient $q!/j!$ is the integer product $(j+1)(j+2)\cdots q$, with the empty product equal to $1$ when $j=q$. Hence $N \in \mathbb{N}$. Also every term defining $x$ is strictly positive, so $x > 0$.
[/step]
[step:Bound the tail by a geometric series]
For every integer $j \ge 1$,
\begin{align*}
\frac{q!}{(q+j)!}
= \frac{1}{(q+1)(q+2)\cdots(q+j)}
\le \frac{1}{(q+1)^j},
\end{align*}
because the denominator contains $j$ factors and each factor is at least $q+1$. Therefore, by comparison with the convergent geometric series of ratio $1/(q+1)$,
\begin{align*}
x
= \sum_{j=1}^{\infty} \frac{q!}{(q+j)!}
\le \sum_{j=1}^{\infty} \frac{1}{(q+1)^j}
= \frac{1/(q+1)}{1 - 1/(q+1)}
= \frac{1}{q}.
\end{align*}
Since $q \ge 2$, it follows that
\begin{align*}
0 < x \le \frac{1}{q} \le \frac{1}{2} < 1.
\end{align*}
[guided]
The tail $x$ is the part of $q!e$ that remains after the terms through $j=q$ have been removed. We estimate each summand in that tail. For an integer $j \ge 1$, cancellation of the factorials gives
\begin{align*}
\frac{q!}{(q+j)!}
= \frac{1}{(q+1)(q+2)\cdots(q+j)}.
\end{align*}
The denominator has exactly $j$ factors, and each factor is at least $q+1$. Hence
\begin{align*}
(q+1)(q+2)\cdots(q+j) \ge (q+1)^j,
\end{align*}
so
\begin{align*}
\frac{q!}{(q+j)!}
\le \frac{1}{(q+1)^j}.
\end{align*}
We now compare the whole tail with the corresponding geometric series. The ratio $1/(q+1)$ lies in $(0,1)$ because $q \in \mathbb{N}$, so the geometric series formula applies:
\begin{align*}
x
= \sum_{j=1}^{\infty} \frac{q!}{(q+j)!}
\le \sum_{j=1}^{\infty} \frac{1}{(q+1)^j}
= \frac{1/(q+1)}{1 - 1/(q+1)}
= \frac{1}{q}.
\end{align*}
The earlier step gave $q \ge 2$, so
\begin{align*}
0 < x \le \frac{1}{q} \le \frac{1}{2} < 1.
\end{align*}
Thus the tail is positive but strictly smaller than one.
[/guided]
[/step]
[step:Derive the contradiction from a non-integer fractional tail]
We have shown that
\begin{align*}
q!e = N + x,
\end{align*}
where $N \in \mathbb{N}$ and $0 < x < 1$. Hence $q!e$ lies strictly between the consecutive integers $N$ and $N+1$, so $q!e \notin \mathbb{N}$. This contradicts the conclusion from the rational assumption that $q!e \in \mathbb{N}$. Therefore $e$ is irrational.
[/step]
