Have I proved in ZF $\aleph (X) \lt \aleph (\mathscr P^3(X))$? As a reminder, I've been a lawyer for more than 25 years after bailing on graduate school.  I recently reacquired the itch to do math.  To that end, I've been (slowly) working through Set Theory:  An Introduction to Independence Proofs by Kunen.  This is Problem 12(a) of Chapter 1.  I've been through the questions here (and on other sites) but I didn't see a sufficiently clear statement of the proof to have confidence that I have this right.  So here goes.
Working in ZF (not assuming AC), we define Hartog's $\aleph$-function via:

$$\aleph(X) = \cup \{ \alpha ~|~ \exists f: \alpha \rightarrow X \text{ with }f \text{ 1-1}\}.$$

To prove: $\aleph(X) \lt \aleph(\mathscr P^3(X))$.
Note:  Kunen remarks that it's easier to prove $\aleph(X) \lt \aleph(\mathscr P^4(X))$.  I'm pretty sure I see that.  Let $R$ be any well-ordering of $Y \subseteq X$.  Then $R \in \mathscr P^3(X)$ and we can define
$$T_\alpha = \{R \in \mathscr P^3(X)~|~\exists Y \subseteq X \text{ such that }R \text{ is a well ordering of } Y \text{ with order type } \alpha \} \in \mathscr P^4(X).$$
$T_\alpha$ is non-empty for any $\alpha \lt \aleph(X)$ because by the definition of $\aleph(X)$ there is a $1-1$ map $f:\alpha \rightarrow X$ which defines a well-order of type $\alpha$ on $f(\alpha) \subseteq X$.  Then $f: \aleph (X) \rightarrow \mathscr P^4(X)$ defined by $f(\alpha) = T_\alpha$ is an injection, proving the inequality.  But that's not the result we were asked to prove.
Resuming:  Let $R$ be any well order of a subset of $X$.  We define a set containing all of the initial segments of $R$:
$$S_R = \{Y \subseteq \text{ dom}(R)~|~Y \text{ is an initial segment of dom}(R) \text{ under } R\} \in \mathscr P^2(X).$$
Define $T(R)$ as the unique ordinal $\alpha$ that maps into $X$ with precisely $S_R$ as its initial segments.
Define $Z(\alpha) = \{S_R~|~T(R) = \alpha \} \in \mathscr P^3(X)$.  Then $Z:\aleph(X) \rightarrow \mathscr P^3(X)$ is $1-1$ and $\aleph(\mathscr P^3(X)) \gt \aleph(X)$ as required.
Is this a correct proof?  Have I elided any steps that aren't (or at least shouldn't be) obvious?  Thanks in advance for any help.
 A: Yes, the proof is fine. But you can somewhat simplify it.

Lemma. If $f\colon A\to B$ is surjective, then there is an injective $g\colon B\to\mathcal P(A)$.

This is easy to prove, in fact you can define $g$ from $\mathcal P(B)$. And it gives us the following corollary.

Corollary. If there is a surjective function $f\colon A\to\aleph(B)$, then $\aleph(B)<\aleph(\mathcal P(A))$.

Now, why do we care about these? Well, because it's easy to define a surjection from $\mathcal P^2(X)$ onto $\aleph(X)$:
$$f(\mathscr C)=\begin{cases}\operatorname{otp}(\mathscr C) & (\mathscr C,\subsetneq)\text{ is well-ordered}\\ 0 & \text{otherwise}\end{cases}$$
This is a surjection, since if $f\alpha\to X$ is injective, it defines a chain of subsets which is isomorphic to $\alpha$ via $f$.

Alternatively, define $T_\alpha$ to be the set of all chains of subsets of $X$ of order type $\alpha$ by $\subsetneq$, so $T_\alpha\in\mathcal P^3(X)$, and $g(\alpha)=T_\alpha$ is an injection from $\aleph(X)$.
(The proof is the same as above.)
It's just a bit cleaner and doesn't require you to parse the domains of relations etc.
