Cardinality of Power Set of Finite Set/Informal Proof

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 $S$.

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.