Given the numbers $ 1,2,2^2, \ldots ,2^{n\minus{}1}$, for a specific permutation $ \sigma \equal{} x_1,x_2, \ldots, x_n$ of these numbers we define $ S_1(\sigma) \equal{} x_1$, $ S_2(\sigma)\equal{}x_1\plus{}x_2$, $ \ldots$ and $ Q(\sigma)\equal{}S_1(\sigma)S_2(\sigma)\cdot \cdot \cdot S_n(\sigma)$. Evaluate $ \sum 1/Q(\sigma)$, where the sum is taken over all possible permutations.
1. **Base Case:**
   For \( n = 1 \), the only number is \( 2^0 = 1 \). The only permutation is \( \sigma = (1) \). Thus, \( S_1(\sigma) = 1 \) and \( Q(\sigma) = S_1(\sigma) = 1 \). Therefore, 
   \[
   \sum_\sigma \frac{1}{Q(\sigma)} = \frac{1}{1} = 1.
   \]
   This matches the expression \( \frac{1}{x_1 \cdot x_2 \cdots x_n} \) since \( x_1 = 1 \).

2. **Inductive Step:**
   Assume the statement is true for \( n = k \), i.e., 
   \[
   \sum_{\sigma \in S_k} \frac{1}{Q(\sigma)} = \frac{1}{x_1 \cdot x_2 \cdots x_k}.
   \]
   We need to show it holds for \( n = k+1 \).

3. **Permutations and Sets:**
   Let \( S(k+1) \) be the set of permutations of \( \{x_1, x_2, \ldots, x_{k+1}\} \) that end in \( x_{k+1} \), and let \( T(k+1) \) be the set of permutations of \( \{x_1, x_2, \ldots, x_k\} \).

4. **Sum Over Permutations:**
   Consider the sum over all permutations:
   \[
   \sum_\sigma \frac{1}{Q(\sigma)} = \sum_{x_{k+1}} \sum_{\sigma \in S(k+1)} \frac{1}{Q(\sigma)}.
   \]
   For each \( x_{k+1} \), the sum can be written as:
   \[
   \sum_{\sigma \in S(k+1)} \frac{1}{Q(\sigma)} = \sum_{\sigma \in T(k+1)} \frac{1}{Q(\sigma) \cdot (x_1 + x_2 + \cdots + x_{k+1})}.
   \]

5. **Applying Inductive Hypothesis:**
   By the inductive hypothesis, we have:
   \[
   \sum_{\sigma \in T(k+1)} \frac{1}{Q(\sigma)} = \frac{1}{x_1 \cdot x_2 \cdots x_k}.
   \]
   Therefore,
   \[
   \sum_{\sigma \in S(k+1)} \frac{1}{Q(\sigma)} = \frac{1}{x_1 \cdot x_2 \cdots x_k \cdot (x_1 + x_2 + \cdots + x_{k+1})}.
   \]

6. **Summing Over All \( x_{k+1} \):**
   Summing over all possible \( x_{k+1} \):
   \[
   \sum_{x_{k+1}} \sum_{\sigma \in S(k+1)} \frac{1}{Q(\sigma)} = \frac{1}{x_1 \cdot x_2 \cdots x_k} \sum_{x_{k+1}} \frac{1}{x_1 + x_2 + \cdots + x_{k+1}}.
   \]
   Since \( x_{k+1} \) can be any of \( x_1, x_2, \ldots, x_{k+1} \), we have:
   \[
   \sum_{x_{k+1}} \frac{1}{x_1 + x_2 + \cdots + x_{k+1}} = \frac{x_1 + x_2 + \cdots + x_{k+1}}{x_1 \cdot x_2 \cdots x_{k+1}}.
   \]

7. **Final Simplification:**
   Therefore,
   \[
   \sum_\sigma \frac{1}{Q(\sigma)} = \frac{1}{x_1 \cdot x_2 \cdots x_k} \cdot \frac{x_1 + x_2 + \cdots + x_{k+1}}{x_1 \cdot x_2 \cdots x_{k+1}} = \frac{1}{x_1 \cdot x_2 \cdots x_{k+1}}.
   \]

By induction, the statement holds for all \( n \).

The final answer is \( \boxed{ \frac{1}{x_1 \cdot x_2 \cdots x_n} } \).