Let $\mathcal{G}(n)$ be the set of graphs on $n$ vertices. Then there is the identity $$ \sum_{\substack{G\in\mathcal{G}(n)\\G\text{ connected}}} (-1)^{|G|} = (-1)^{n-1} (n-1)!, $$ where the sum is over graphs that are connected. This has been discussed for example in this question.
I am curious about the fact that the right-hand side is also equal to the formula for the Möbius function $\mu_{\mathcal{P}(n)}(0,1)$, where $\mathcal{P}(n)$ is the poset of partitions of an $n$-element set, and $0$ and $1$ are the minimal and maximal partitions.
I don't think this a total coincidence for the reason that there is an order-preserving mapping $f:\mathcal{G}(n)\to\mathcal{P}(n)$ which sends $G$ to the partition of $[n]$ into connected components of $G$. Here we make $\mathcal{G}(n)$ into a poset by saying $G\leq G'$ if the edge set of $G$ is a subset of the edge set of $G'$. Then $\mathcal{G}(n)$ is isomorphic to the boolean algebra on $\binom{n}{2}$ elements, with Möbius function $\mu_{\mathcal{G}(n)}(H,G) = (-1)^{|G\setminus H|}$.
With this notation we can rewrite the identity above in the following suggestive way: $$ \sum_{H\in f^{-1}(0)} \sum_{G\in f^{-1}(1)} \mu_{\mathcal{G}(n)}(H,G) = \mu_{\mathcal{P}(n)}(0,1). $$
Now there is an identity due to Baclawski which states that if $f:P\to Q$ is an order-preserving mapping between posets then $$ \mu(Q) = \mu(P) + \sum_{y\in Q} \mu(f^{-1}(Q_{\leq y})) \mu(Q_{>y}). $$ Here the Möbius number of a poset $P$ is defined by $\mu(P)=\mu_{\hat{P}}(\hat{0},\hat{1})$, and $\hat{P} = P\cup\{\hat{0},\hat{1}\}$ is a new poset formed by adding a minimal element $\hat{0}$ and a maximal element $\hat{1}$.
I am curious whether there is some way to use this identity to prove this result for the signed sum of graphs.