Let $f(a,b,c)$ be the number of ways of writing a set of size $a$ as a union of $b$ distinct subsets of size $c$. I've noticed that $$\sum_{b=0}^{\binom{a}{c}}(-1)^{b+1}f(a,b,c)=(-1)^{a+c}\binom{a-1}{c-1}\tag{*}$$ This can be proved as follows. According to this answer, $f$ is given by $$f(a,b,c) = \sum_{i=0}^a(-1)^i\binom{a}{i}\binom{\binom{a-i}{c}}{b}$$ Then, $$\sum_{b=0}^{\binom{a}{c}}(-1)^{b+1}f(a,b,c)=\sum_{b=0}^{\binom{a}{c}}(-1)^{b+1}\sum_{i=0}^{a}(-1)^i\binom{a}{i}\binom{\binom{a-i}{c}}{b} = \sum_{i=0}^{a}(-1)^i\binom{a}{i}\sum_{b=0}^{\binom{a}{c}}(-1)^{b+1}\binom{\binom{a-i}{c}}{b}. \tag{**}$$
Now by the binomial theorem, $$\sum_{i=0}^n(-1)^i\binom{n}{i}=\begin{cases}0 & n > 0\\ 1 & n =0 \end{cases}.$$ Using this identity and the fact that $\binom{n}{k}=0$ for $k>n$, the inner sum of $(**)$ can be written as $$\sum_{b=0}^{\binom{a}{c}}(-1)^{b+1}\binom{\binom{a-i}{c}}{b}=\begin{cases} 0 & i\leq a-c\\ -1 & \text{otherwise} \end{cases}.$$ Substituting into $(**)$, $$\sum_{i=0}^{a}(-1)^i\binom{a}{i}\sum_{b=0}^{\binom{a}{c}}(-1)^{b+1}\binom{\binom{a-i}{c}}{b} = \sum_{i=a-c+1}^{a}(-1)^{i+1}\binom{a}{i} = \sum_{i=0}^{c-1}(-1)^{i+a+1}\binom{a}{i}.$$ Lastly, using the formula for alternating partial sums of a row in pascal's triangle, $$\sum_{i=0}^{c-1}(-1)^{i+a+1}\binom{a}{i}=(-1)^{a+c}\binom{a-1}{c-1}.$$
This clean closed form makes me wonder if there is a simpler way of proving $(*)$. In particular, is there a simple counting argument that can prove $(*)$ without computing $f(a,b,c)$?
The alternating nature of the sum in $(*)$ suggests it may be a form of the inclusion-exclusion principle, but I have not been able to model it as such.
EDIT
One potential way forward is to use ideas from this answer. Fix $a$ and $c$. Let $\mathcal{R}=\{S\subseteq[a]:|S|=c\}$. Let $I:\mathcal{P}([a])\rightarrow \mathcal{P}([a])$ be the indicator function for covers of $[a]$. Then, $$\sum_{b=0}^{\binom{a}{c}}(-1)^{b}f(a,b,c)=\sum_{\mathcal{S}\subseteq \mathcal{R}}(-1)^{|\mathcal{S}|}I(\mathcal{S})$$ so it would remain to show that the RHS is $(-1)^{a+c+1}\binom{a-1}{c-1}$. The form on the RHS is also promising because it resembles the mobius inversion of $I$ on the boolean lattice $B_a$. In particular, if we define $$g(\mathcal{T})=\sum_{\mathcal{S}\subseteq \mathcal{T}}(-1)^{|\mathcal{T}|-|\mathcal{S}|}I(\mathcal{S})$$ then $g$ is the mobius inversion of $I$, or equivalently, $$I(\mathcal{T}) = \sum_{\mathcal{S}\subseteq \mathcal{T}}g(\mathcal{S})$$ Thus, if there is a "nice" formula for $g$, then calculating $g(\mathcal{R})=(-1)^{\binom{a}{c}+a+c+1}\binom{a-1}{c-1}$ would prove $(*)$. The hope is that such a proof would translate to a nice counting argument without the language of boolean lattices.
