设事件 $\omega_n$ 表示随机取 $n$ 个 $(0, 1)$ 中的实数, 其和小于 $1$.
$$\begin{aligned}
E &= \sum_{n = 1}^{\infty} n P(\omega_{n - 1} \overline{\omega_n}) \\
&= \sum_{n = 1}^{\infty} n \left[ P(\omega_{n - 1}) - P(\omega_{n - 1}\omega_n) \right] \\
&= \sum_{n = 1}^{\infty} n \left[ P(\omega_{n - 1}) - P(\omega_n) \right] \\
&= 1 + \sum_{n = 1}^{\infty} P(\omega_n)
\end{aligned}
$$
考虑根据定义求 $\omega_n$:
$$\omega_n = \int_{0}^1 \mathrm{d}x_1 \int_{0}^{1 - x_1} \mathrm{d}x_2 \int_{0}^{1 - x_1 -x_2} \mathrm{d}x_3 \cdots \int_{0}^{1 - \sum_{i < n} x_i} \mathrm{d}x_n
$$
将 $1$ 用形式变量 $x$ 代替, 即得
$$\omega_n(x) = \int_{0}^x \mathrm{d}x_1 \int_{0}^{x - x_1} \mathrm{d}x_2 \int_{0}^{x - x_1 -x_2} \mathrm{d}x_3 \cdots \int_{0}^{x - \sum_{i < n} x_i} \mathrm{d}x_n
$$
考虑寻找 $\omega_{n + 1}$ 和 $\omega_n$ 之间的关系, 有
$$\omega_{n + 1}(x) = \int_{0}^{x} \omega_n(x - x_{n + 1}) \mathrm{d}x_{n + 1}
$$
注意到在积分内部, $\mathrm{d}(x - x_{n + 1})$ 与 $\mathrm{d}x_{n + 1}$ 实际上是等价的, 于是得到递推式
$$\omega_{n + 1}(x) = \int_{0}^{x} \omega_n(t) \mathrm{d}t
$$
根据定义 $\omega_1(x) = x$, 于是归纳得
$$\omega_{n}(x) = \int_{0}^{x} \frac{t^{n - 1}}{(n - 1)!} \mathrm{d}t = \frac{x^n}{n!}
$$
将 $x$ 换回 $1$, 回代得
$$E = 1 + \sum_{n = 1}^{\infty} \frac{1}{n!} = \mathrm{e}
$$