Induction can prove it. Here is a simple proof:
Let f (a) = the product of elements of a, and the sum of the product of elements of a subset of g (a) = a.
Then the subset of A {x_(n+ 1)} is divided into two types: X and X {x_(n+ 1)}, where X is a subset of A. If X is not empty, then f(X and {x _ (n+1)}). If x is empty, then f (x and {x _ (n+ 1)}) = x _ So g(A and {x _ (n+1)}) = x _ (n+1)+g (a)+x.