Cardinality of Power Set of Finite Set/Proof 2

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
We can see that enumerating the subsets of $S$ is equivalent to counting all of the ways of selecting $k$ out of the $n$ elements of $S$ with $k = 0, 1, \ldots, n$.

So, from Cardinality of Set of Subsets, the number we are looking for is:


 * $\displaystyle \left|{\mathcal P \left({S}\right)}\right| = \sum_{k=0}^n \binom n k$

But from the binomial theorem:


 * $\displaystyle \left({x + y}\right)^n = \sum_{k=0}^n \binom n k x^{n-k} y^k$

It follows that:
 * $2^n = \displaystyle \left({1 + 1}\right)^n = \sum_{k=0}^n \binom n k \left({1}\right)^{n-k} \left({1}\right)^k = \sum_{k=0}^n        \binom n k = \left|{\mathcal P \left({S}\right)}\right|$