Cantor's Theorem (Strong Version)

Theorem
Let $S$ be a set.

Let $\mathcal P^n \left({S}\right)$ be defined recursively by:
 * $\mathcal P^n \left({S}\right) = \begin{cases}

S & : n = 0 \\ \mathcal P \left({\mathcal P^{n-1} \left({S}\right)}\right) & : n > 0 \end{cases}$ where $\mathcal P \left({S}\right)$ denotes the power set of $S$.

Then $S$ is not equivalent to $\mathcal P^n \left({S}\right)$ for any $n > 0$.

Proof
We temporarily introduce the notation:
 * $a^n = \begin{cases}

a : & n = 0 \\ \left\{{a^{n-1}}\right\} : & n > 0 \end{cases}$ where $a \in S$.

Thus $a^n$ consists of a single element $a^{n-1} \in \mathcal P^{n-1} \left({S}\right)$.

Let there be a bijection $f: S \to \mathcal Q^n$ where $\mathcal Q^n \subseteq \mathcal P^n \left({S}\right)$.

Then define:
 * $\mathcal A^{n-1} = \left\{{s^{n-1} \in \mathcal P^{n-1} \left({S}\right): s^{n-1} \ne f \left({s}\right)}\right\}$

Each $\mathcal Q^{n-1} \in \mathcal Q^n$ is the image of some element of $S$ under $f$.

Let $\mathcal Q^{n-1} = f \left({x}\right)$ for some $x \in S$.

Let $x^{n-1} \in \mathcal Q^{n-1} = f \left({x}\right)$.

Then $x^{n-1} \notin \mathcal A^{n-1}$, and so $ A^{n-1} \notin \mathcal Q^{n-1}$.

On the other hand, let $x^{n-1} \notin \mathcal Q^{n-1} = f \left({x}\right)$.

Then $x^{n-1} \in \mathcal A^{n-1}$, and so again $ A^{n-1} \notin \mathcal Q^{n-1}$.

Thus $\mathcal A^{n-1} \notin Q^n$.

Therefore $Q^n$ is a proper subset of $\mathcal P^n \left({S}\right)$.

Hence the result.