This question is motivated by one of the problems from this year's Putnam Exam set. That is,

>**Problem.** 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
$$M_{ij} = \begin{cases} 0 & \mbox{if }S_i \cap S_j = \emptyset; \\
1 & \mbox{otherwise.}
\end{cases}.$$
Calculate the determinant of $M$. **Answer:** $-1$.

I like to consider the following variant which got me puzzled.

> **Question.** Preserve the notation from above, let
$A$ be the matrix whose $(i,j)$ entry is
$$A_{ij} = \begin{cases} 1 & \# (S_i \cap S_j) =1; \\
0 & \mbox{otherwise.}
\end{cases}.$$
If $n>1$, depending (?) on the ordering of the subsets, is this true?
$$\det(A)=\pm\prod_{k=1}^nk^{\binom{n}k}.$$

*Remark.* Amusingly, the same number counts "product of sizes of all the nonempty subsets of $[n]$" according to [OEIS][1].

[1]: https://oeis.org/A229333