Let $S_1, S_2, \dots, S_{2^n - 1}$ be the nonempty subsets of $\{1, 2, \dots, n\}$ in some order, and let $M$ be the $(2^n - 1) \times (2^n - 1)$ matrix whose $(i, j)$ entry is
\[m_{ij} = \left\{
\begin{array}{cl}
0 & \text{if $S_i \cap S_j = \emptyset$}, \\
1 & \text{otherwise}.
\end{array}
\right.\]
Calculate the determinant of $M$.
1. **Define the Matrix \( M \)**:
   Let \( S_1, S_2, \dots, S_{2^n - 1} \) be the nonempty subsets of \( \{1, 2, \dots, n\} \). The matrix \( M \) is defined such that its \((i, j)\)-th entry is:
   \[
   m_{ij} = \left\{
   \begin{array}{cl}
   0 & \text{if } S_i \cap S_j = \emptyset, \\
   1 & \text{otherwise}.
   \end{array}
   \right.
   \]

2. **Reformulate the Problem**:
   We can reformulate the problem using the Iverson bracket notation:
   \[
   m_{ij} = \left[ S_i \cap S_j \neq \emptyset \right].
   \]
   We need to calculate the determinant of \( M \).

3. **Introduce the Lemma**:
   We use the following lemma to express \( m_{ij} \):
   \[
   \left[ S_i \cap S_j \neq \emptyset \right] = \sum_{U \in P} (-1)^{|U| - 1} \left[ U \subseteq S_i \right] \left[ U \subseteq S_j \right],
   \]
   where \( P \) is the set of all nonempty subsets of \( \{1, 2, \dots, n\} \).

4. **Define Matrices \( B \) and \( C \)**:
   Define two \( P \times P \) matrices:
   \[
   B = \left( (-1)^{|T| - 1} \left[ T \subseteq S \right] \right)_{(S, T) \in P \times P},
   \]
   \[
   C = \left( \left[ S \subseteq T \right] \right)_{(S, T) \in P \times P}.
   \]

5. **Matrix Decomposition**:
   Using the lemma, we can write:
   \[
   M = BC.
   \]
   Therefore, the determinant of \( M \) is:
   \[
   \det M = \det(B) \cdot \det(C).
   \]

6. **Calculate \( \det(B) \)**:
   We order \( P \) such that if \( S \subseteq T \), then \( S \leq T \). This makes \( B \) a lower-triangular matrix. The determinant of a lower-triangular matrix is the product of its diagonal entries:
   \[
   \det B = \prod_{S \in P} (-1)^{|S| - 1} = (-1)^{\sum_{S \in P} (|S| - 1)}.
   \]
   We need to calculate \( \sum_{S \in P} (|S| - 1) \). Each element of \( \{1, 2, \dots, n\} \) appears in \( 2^{n-1} \) subsets, so:
   \[
   \sum_{S \in P} |S| = n \cdot 2^{n-1}.
   \]
   There are \( 2^n - 1 \) nonempty subsets, so:
   \[
   \sum_{S \in P} (|S| - 1) = n \cdot 2^{n-1} - (2^n - 1) = n \cdot 2^{n-1} - 2^n + 1.
   \]
   This simplifies to:
   \[
   n \cdot 2^{n-1} - 2^n + 1 \equiv [n \neq 1] \pmod{2}.
   \]
   Therefore:
   \[
   \det B = (-1)^{[n \neq 1]}.
   \]

7. **Calculate \( \det(C) \)**:
   The matrix \( C \) is upper-triangular with all diagonal entries equal to 1:
   \[
   \det C = 1.
   \]

8. **Combine Results**:
   Combining the results, we get:
   \[
   \det M = \det(B) \cdot \det(C) = (-1)^{[n \neq 1]} \cdot 1 = (-1)^{[n \neq 1]}.
   \]

The final answer is \( \boxed{ (-1)^{[n \neq 1]} } \).