Consider some list $A=(a_1,a_2,\cdots,a_n)$. I'd like to find a closed form for the following operation.
$$f(A)=\sum_{k=1}^n(-1)^{k-1}s_k= s_1-s_2+\cdots(-1)^{n-1}s_n.$$ Where $s_k$ is the sum of all combinations of products of $k$ unique elements of $A$. For example, if $A=(x,y,z)$, then $$s_1=x+y+z,$$ $$s_2=xy+xz+yz,$$ $$s_3=xyz.$$ Thus $f(A)=(x+y+z)-(xy+xz+yz)+(xyz)$.
What is the "nicest" way to formulate this sum of products?
Is this operation known/common in combinatorics?
The best I can come up with is $$\sum_{k=1}^n(-1)^{k-1}\prod_{a\in\mathcal{P}_n(A)}a$$ where $\mathcal{P}_n(A)$ denotes the set of all subsets in $A$ with cardinality $n$. However, it would be nice to have this in some "standard form" without the use of powersets (let alone the generalization of such). My binomial theorem alarm is going off; thus I imagine there is some nicer closed form which makes use of binomial coefficients.
Any insight or advice would be greatly appreciated.
What you got is correct. We directly use the binomial expansion to get \begin{align} \prod_{k =1}^n (1-a_k) &= \sum_{j = 0}^n \sum_{C \in \binom{[n]}{j}} \prod_{c \in C} -a_c \\ &= \sum_{j = 0}^n \sum_{C \in \binom{[n]}{j}} (-1)^{j} \prod_{c \in C} a_c \\ &= 1 + \sum_{j = 1}^n \sum_{C \in \binom{[n]}{j}} (-1)^{j} \prod_{c \in C} a_c \\ &= 1 + \sum_{j = 1}^n (-1)^{j} \sum_{C \in \binom{[n]}{j}} \prod_{c \in C} a_c \\ &= 1 - \sum_{j = 1}^n (-1)^{j - 1} \sum_{C \in \binom{[n]}{j}} \prod_{c \in C} a_c \\ &= 1 - \sum_{j = 1}^n (-1)^{j - 1} s_j \end{align}