I think it is important to consider the context in which this question was posed.
We first have Cantor's Theorem (Theorem 5, p.32) which says that $| A | < | \mathcal{P} (A) |$ for any set $A$.
But we also have Theorem 7 (p.34), which I will quote in its entirety:
Let $\mathcal{M}$ be a set of cardinals such that $\forall \mathfrak{m} \in \mathcal{M} \exists \mathfrak{n} \in \mathcal{M} ( \mathfrak{m} < \mathfrak{n} )$, and let $\mathcal{A}$ be a family of sets such that $\forall \mathfrak{m} \in \mathcal{M} \exists A \in \mathcal{A} ( |A| = \mathfrak{m} )$. Then $| \bigcup \mathcal{A} |$ exceeds every cardinal in $\mathcal{M}$.
It is this last ingredient that allows this process to extend beyond the simple successor cases. (And this is what everyone else has been saying.)
Start with $X_0 , X_1 = \mathcal{P}(X_0) , X_2 = \mathcal{P} (X_1) , \ldots , X_{n+1} = \mathcal{P} ( X_n ) , \ldots $ and then after $\omega$ steps take $X_\omega = \bigcup_{n < \omega} X_n$. By Theorem $7$ we know that $| X_\omega | > | X_n |$ for all $n < \omega$. We continue with $X_{\omega + 1} , X_{\omega_2} , \ldots , X_{\omega \cdot 2} , X_{\omega \cdot 3} , \ldots , X_{\omega \cdot \omega}$ etc. We can even imagine this extending beyond the countable ordinals, arriving at $X_{\omega_1}$ which will satisfy $| X_{\omega_1} | > | X_\alpha |$ for all $\alpha < \omega_1$. And this can continue still more.