Example 4 Given the numbers $1,2,2^{2}, \cdots, 2^{n-1}$. For any permutation $\sigma=\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ of them, define $S_{1}(\sigma)=$ $x_{1}, S_{2}(\sigma)=x_{1}+x_{2}, \cdots, S_{n}(\sigma)=x_{1}+x_{2}+\cdots+$ $x_{n}$. Also let $Q(\sigma)=S_{1}(\sigma) S_{2}(\sigma) \cdots S_{n}(\sigma)$, find $\sum \frac{1}{Q(\sigma)}$. (The sum is taken over all permutations)
(1989, Canadian Mathematical Competition)
Analysis: Generalize, first calculate the value of $S\left(a_{1}, a_{2}, \cdots, a_{n}\right)=\sum \frac{1}{Q(\sigma)}$ for any $n$ positive numbers $a_{1}, a_{2}, \cdots, a_{n}$.
When $n=1$, $S\left(a_{1}\right)=\frac{1}{a_{1}}$.
When $n=2$,
$$
\begin{aligned}
S\left(a_{1}, a_{2}\right) & =\frac{1}{a_{1}\left(a_{1}+a_{2}\right)}+\frac{1}{a_{2}\left(a_{1}+a_{2}\right)} \\
& =\frac{1}{a_{1} a_{2}} .
\end{aligned}
$$

When $n=3$,
$$
\begin{array}{c}
S\left(a_{1}, a_{2}, a_{3}\right)=\frac{1}{a_{1}\left(a_{1}+a_{2}\right)\left(a_{1}+a_{2}+a_{3}\right)} \\
+\frac{1}{a_{1}\left(a_{1}+a_{3}\right)\left(a_{1}+a_{2}+a_{3}\right)} \\
+\frac{1}{a_{2}\left(a_{2}+a_{1}\right)\left(a_{1}+a_{2}+a_{3}\right)} \\
+\frac{1}{a_{2}\left(a_{2}+a_{3}\right)\left(a_{1}+a_{2}+a_{3}\right)} \\
\quad+\frac{1}{a_{3}\left(a_{3}+a_{1}\right)\left(a_{1}+a_{2}+a_{3}\right)} \\
\quad+\frac{1}{a_{3}\left(a_{3}+a_{2}\right)\left(a_{1}+a_{2}+a_{3}\right)} \\
=\frac{1}{\left(a_{1}+a_{2}+a_{3}\right)}\left[\frac{1}{a_{2}+a_{1}}\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}\right)\right. \\
\left.+\frac{1}{a_{1}+a_{3}}\left(\frac{1}{a_{1}}+\frac{1}{a_{3}}\right)+\frac{1}{a_{2}+a_{3}}\left(\frac{1}{a_{2}}+\frac{1}{a_{3}}\right)\right] \\
=\frac{1}{a_{1} a_{2} a_{3}} .
\end{array}
$$

Induction from the above special cases, conjecture that $S\left(a_{1}, a_{2}, \cdots, a_{n}\right)=\frac{1}{a_{1} a_{1} \cdots a_{n}}$.

Proof: Assume the conclusion holds for $n=k$. Then, for $n=k+1$, according to the calculation method for $n=3$, the $(k+1)!$ permutations of $k+1$ numbers $a_{1}, a_{2}, \cdots, a_{k+1}$ can be divided into $k+1$ classes. Each class contains $k!$ permutations with the same element $a_{i}$ as the first element. The last factor in these permutations is always $S\left(=a_{1}+\right.$ $\left.a_{2}+\cdots+a_{k+1}\right)$, and the first $k$ factors are all possible permutations of the remaining $k$ elements. By the induction hypothesis, the sum of $\frac{1}{Q(\sigma)}$ for these $k!$ permutations is
$$
\frac{1}{a_{1} a_{2} \cdots a_{i-1} a_{i+1} \cdots a_{k+1}} \cdot \frac{1}{S}=\frac{a_{i}}{a_{1} a_{2} \cdots a_{k+1} S} .
$$

Thus, $S\left(a_{1}, a_{2}, \cdots, a_{k+1}\right)=\frac{1}{a_{1} a_{2} \cdots a_{k+1} S} \sum_{i=1}^{k+1} a_{i}$
$$
=\frac{1}{a_{1} a_{2} \cdots a_{k+1}} .
$$

Therefore, the result we are looking for is
$$
\sum \frac{1}{Q(\sigma)}=\frac{1}{1 \cdot 2 \cdot 2^{2} \cdots 2^{n-1}}=2^{-\frac{n(n-1)}{2}} .
$$