Suppose that $\mathcal{F}$ is an exponential family in the combinatorial sense (see these slides or Chapter 3 of generatingfunctionology). Let $h(n, k)$ be the number of hands of weight $n$ with exactly $k$ cards. Given a vector $v \in \{-1, 1\}^n$, define the following:
\begin{align*} \operatorname{sgn} v &= (-1)^{\text{number of $-1$'s in $v$}} \\ w(v) &= \sum_{k = 1}^n v_k. \end{align*}
I believe that I have proven the following identity with generating functions:
$$\sum_{\substack{v \in \{-1, 1\}^n \\ k}} (\operatorname{sgn} v) h(n, k) w(v)^k = n! (2d)^n,$$
where $d$ is the number of cards in the deck of weight $1$ of $\mathcal{F}$. (This is a finite sum since $h(n, k) = 0$ if $k < 0$ or $k > n$.)
This feels too general to be true. For example, the above would hold when $h(n, k)$ is the number of:
- Labeled graphs on $n$ vertices with $k$ connected components;
- Partitions of $\{1, \dots, n\}$ into $k$ disjoint, nonempty parts; or
- Permutations of $\{1, \dots, n\}$ with $k$ disjoint cycles.
I've numerically checked the last two examples (these are the Stirling numbers of the second and first kind, respectively), and they seem to work.
Is there a combinatorial proof of this identity?