TITLE: Is there an intuitive proof of the identity $ \sum_{L \subset S} \prod_{x \in L} (x-1) = \prod_{x \in S} x$ from general principles?
QUESTION [10 upvotes]: Let $R$ be a commutative ring with unity, and let $S\subset R$ be any finite set.  Then
$$ \sum_{L \subset S} \prod_{x \in L} (x-1) = \prod_{x \in S} x,$$
which is easy enough to show by induction.
Does this follow from any sort of general principle?  Perhaps inclusion-exclusion, in some form?

REPLY [9 votes]: Let $S'$ be $\{x : x-1 \in S\}$, in other words we shift $S$ by one.
The equality
$$ \sum_{L \subset S} \prod_{x \in L} (x-1) = \prod_{x \in S} x$$
reduces to
$$ \sum_{L \subset S'} \prod_{x \in L} x = \prod_{x \in S'} (x+1)$$
but this is obviously distributivity: LHS is the expanded RHS.