= Solution
The <exponential series> at $1$ gives
$$
\boxed{e=\sum_{n=0}^{\infty}\frac1{n!}.}
$$
For every positive <integer> $q$, multiply by the <factorial> $q!$ and separate the integer part:
$$
q!e=\underbrace{\sum_{n=0}^{q}\frac{q!}{n!}}_{I_q\in\mathbb Z}
+R_q,\qquad
R_q=\sum_{j=1}^{\infty}\frac1{(q+1)(q+2)\cdots(q+j)}.
$$
Every term in $R_q$ is positive. Each denominator is at least $(q+1)^j$, and the inequality is strict when $j\ge2$. Comparison with the <geometric series> therefore gives
$$
0<R_q<\sum_{j=1}^{\infty}(q+1)^{-j}=\frac1q\le1.
$$
If $e=p/q$ with integers $p$ and $q\ge1$, then $q!e=p(q-1)!$ would be an <integer>. Since $I_q$ is also an <integer>, $R_q$ would be an <integer> strictly between zero and one, which is impossible. Hence
$$
\boxed{e\text{ is irrational}.}
$$
This proves the <irrationality of e> by using its rapidly decreasing series tail.
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