Use a proof by induction to show that for any finite set S with |S| = n, where n is a nonnegative integer, then Shas 2n subsets.

Sure, here is a proof by induction for the statement:

Base Case (n=0): If a set S has no elements (|S|=0), then it has only one subset, the empty set. So, 2^0 = 1 which is true.

Inductive Step: Assume that a set S with |S|=k has 2^k subsets (this is the inductive hypothesis). We need to show that a set S' with |S'|=k+1 has 2^(k+1) subsets.

Consider a set S' with |S'|=k+1. We can write S' as the union of two sets: S and {a}, where |S|=k and a is an element not in S. The subsets of S' are either subsets of S, or they include a and the rest of the elements form a subset of S.

By the inductive hypothesis, S has 2^k subsets. Each of these can form a subset of S' by adding a, giving another 2^k subsets. Therefore, S' has 2^k + 2^k = 2*2^k = 2^(k+1) subsets.

Therefore, by mathematical induction, the statement is true for all nonnegative integers n.