Cardinality of Power Set of Finite Set

Theorem
Let $$S$$ be a set such that $$\left|{S}\right| = n$$.

Then $$\left|{\mathcal{P} \left({S}\right)}\right| = 2^n$$.

It can be seen that the power set's alternative notation $2^S$ is indeed appropriate.

However, because of possible confusion over the conventional meaning of $$2^n$$, its use is deprecated.

Proof
Let $$T = \left\{{0, 1}\right\}$$.

Let $$\chi: S \to T$$ be defined as:

$$\forall A \subseteq S: \chi \left({A}\right) = \chi_A$$ where:

$$ \forall x \in S: \chi_A \left({x}\right) = \begin{cases} 1 & : x \in A \\ 0 & : x \notin A \end{cases} $$

It is clear that $$\chi: A \to \chi_A$$ is a bijection from $$\mathcal{P} \left({S}\right) \to T^S$$.

Thus by Cardinality of Set of All Mappings the result follows.