Why?

Write \\(S = \lbrace x_1, \ldots, x_n \rbrace\\).

Given this ordering for the elements, each subset \\(A\\) corresponds to a vector of \\(n\\) bits, with the ith bit saying whether or not \\(x_i\\) belongs to \\(A\\).

So each subset can be identified by an n-digit sequence in base 2.

Since there are \\(2^n\\) of these sequences, that means the power set has size \\(2^n\\).