**Proof**: Let $n\in\mathbb{N}.$ Let $\mathcal{B}_{n}$ the set binary string of length $n$ (that is, [[String|strings]] of length $3$ whose elements taken from $\{ 0,1 \}$): $\mathcal{B}_{n}=\{ (c_{i})_{i=0}^{n-1} \mid c_{i} \in \{ 0,1 \} \}$ Note $[\![n]\!]= \{ k\in \mathbb{N} \mid k<n \}.$ Define the function $\begin{align} f :&P([\![n]\!]) \to \mathcal{B}_{n} \\ &A \mapsto (c_{i})_{i=0}^{n} \text{ where } c_{i}= \begin{cases} 1 & i\in A \\ 0 & \text{otherwise.} \end{cases} \end{align}$and $\begin{align} g:&\mathcal{B}_{n} \to P([\![ n]\!]) \\ & (c)_{i=0}^{n-1} \mapsto \{ i \in [\![n]\!] \mid c_{i} = 1 \} \end{align}$Now for all $A\in P([\![n]\!]),$ $g\circ f (A)=A$ by definition. Also for all $x\in \mathcal{B}_{n},$ $f\circ g(x)=x.$ Thus $f$ is bijective. Furthermore, $|\mathcal{B}_{n}|=2^{n}$ since there are $2$ options for each $c_{i}$ in $(c_{i})_{i=0}^{n-1}.$ Thus $|P([\![n]\!])|=|\mathcal{B}_{n}|=2^{n}$ since there exists a bijection between the sets.