Cardinality of Power Set of Finite Set

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

where $\left|{S}\right|$ denotes the cardinality of $S$,

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

where $\mathcal P \left({S}\right)$ denotes the power set of $S$.

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.

Informal Proof
Given an element $x$ of $S$, each subset of $S$ either includes $x$ or does not include $x$ (this follows directly from the definition of a set), which gives us two possibilities. The same reasoning holds for any element of the set.

One can intuitively see that this means that there are $\displaystyle\underbrace{2 \times 2 \times \ldots \times 2}_{\vert S \vert} = 2^{\vert S \vert}$ total possible combinations of elements of $S$, which is exactly $\vert\mathcal{P}(S)\vert$.

Note
The formal mathematical backing for the intuitive leap made in this "proof" is non-trivial, so while this it serves as an excellent demonstration of why this result holds true, it does not constitute a fully rigorous proof of this theorem.

Special Case
This formula even works when $S = \varnothing$.

Clearly:
 * $\mathcal P \left({\varnothing}\right) = \left\{{\varnothing}\right\}$

has one element, that is, $\varnothing$.

So:
 * $\left|{\mathcal P \left({\varnothing}\right)}\right| = \left|{\left\{{\varnothing}\right\}}\right| = 1 = 2^{0}$

thus confirming that the main result still holds when $\left|{S}\right| = 0$.