Proof that axiom of replacement won't generate a set going arbitrarily high in the cumulative hierarchy, without relying on replacement? Is it possible to prove, in first-order ZFC, that the axiom of replacement can't generate a set that includes members going arbitrarily high in the cumulative hierarchy, but without using the axiom of replacement itself for the proof?  Or in other words, to prove without using replacement that the kind of mapping used by replacement won't include elements in its output image going arbitrarily high in the hierarchy?
If it's helpful, part of the motivation for this question is the issue raised here [1] arguing that the replacement axiom doesn't follow directly from the iterative conception of sets: "When Replacement has been justified according to the iterative conception, the reasoning has in fact been circular as it was in Zermelo [1930a], with some feature of the cumulative hierarchy picture newly adduced solely for this purpose."  In particular, one concern you might have with the axiom of replacement, from the standpoint of the iterative conception of sets, is that the mapping in replacement might map to sets arbitrarily high in the hierarchy, so there might be no valid set formed by the image of the mapping (just a class).  Now, if you have full ZFC including replacement, then you can show that the mappings used by replacement can never map arbitrarily high in the hierarchy, but what I'm unclear on is whether this can be shown without relying on the axiom of replacement for any part of the proof (since in the context of the linked paper the goal would be to intuitively justify replacement from the iterative conception, and avoid circularity in this justification).
Here is a related question, but for 2nd order ZFC [2]. Note, the answer to this previous question assumes that there is an inaccessible cardinal in the metatheory, but the current question should rely only on the ZFC axioms.
Edit: remove proof sketch for now, since not sure if it was on the right track.
[1] https://www.jstor.org/stable/41472440
[2] https://math.stackexchange.com/questions/1263848/are-categorical-second-order-axiomatizations-of-set-theory-inconsistent-due-to-t]]
[3] Proof that we can't get $\omega + \omega$ without Replacement
 A: No such proof can exist: a counterexample is given by one of the standard examples of a failure of replacement, $V_{\omega + \omega}$, with the function-class $\{(n,\omega+n)\ |\ n \in \omega\}$.
$V_{\omega+\omega}$ is a model of (ZFC–replacement); moreover, it has an exhaustive cumulative hierarchy, i.e. it believes “every set is in some $V_\alpha$”.  The given function-class is provably a function-class in (ZF–replacement), and its domain is a set in $V_{\omega+\omega}$, but its image goes arbitrarily high within the cumulative hierarchy there.
A: The replacement axiom is equivalent (over the other axioms of ZF) to the assertion that in every instance, the ranks of witnesses are bounded in the ordinals. Specifically, replacement is equivalent to the following principle:
Bounded witness rank axiom. If $A$ is a set and every element $a\in A$ has a unique witness $b$ for some property $\varphi(a,b,c)$, possibly with some fixed parameter $c$, then the ranks of the witnesses $b$ are bounded.
Certainly the replacement axiom implies the bounded witness rank axiom, since if the can be collected into a set, then the rank of that collecting set will be a bound on the ranks of the witnesses themselves.
Conversely, however, if you have the bounded witness rank axiom, then replacement holds, since if you can bound the ranks of the witnesses, then you have found a collecting set for them. Namely, all witnesses appear in some sufficiently large $V_\alpha$, and then you can pick them out exactly as a subset using the separation axiom.
