Another proof is by induction on n.

The base case is for n = 0. That means S = {}. So what must be shown here is that:

\\[|2^{\lbrace \rbrace}| = 2^{|\lbrace \rbrace|} \\]

Well \\(2^{\lbrace \rbrace\}\\) is the set of subsets of the empty set, which just contains one member -- the empty set. So the left hand side of this equation is 1.

On the right hand side, since \\(|\lbrace \rbrace|\\) equals 0, we also get the value 1.

The base case is for n = 0. That means S = {}. So what must be shown here is that:

\\[|2^{\lbrace \rbrace}| = 2^{|\lbrace \rbrace|} \\]

Well \\(2^{\lbrace \rbrace\}\\) is the set of subsets of the empty set, which just contains one member -- the empty set. So the left hand side of this equation is 1.

On the right hand side, since \\(|\lbrace \rbrace|\\) equals 0, we also get the value 1.