Let ω (or all natural numbers) be the set defined by the axiom of infinity in ZF system, ordinal be "transitive and well ordered by ∈", and a set is transitive if all its elements are also its subset.
It seems that induction is now extended to base on ordinal and could be proved, so it's no longer an axiom. ω is considered the smallest part of ordinals, this needs to be proved. And to prove the old induction based on integer, it's also necessary to prove that ω is an ordinal, of cause without using the old axiom of induction. Is it possible ?
To prove directly ω is transitive, it's necessary to prove that all elements of ω are subset of ω. We know that Ø is subset of ω, and for any element n of ω, if n is subset of ω then its successor n⋃{n} is a subset of ω. But from this to "all element of ω is subset of ω", we need old axiom of induction.
If in another way, to prove that ω is an initial segment of ordinals, we need to prove that all elements of ω are ordinal. We know that Ø is an ordinal, and for any element n of ω, if n is an ordinal then its successor n⋃{n} is an ordinal. But from this to prove "all elements of ω are ordinal", we need also the old axiom of induction.
How to prove that ω is an ordinal without using the old axiom of induction ?
We don't use the axiom of induction to prove that $\omega$ is an ordinal. We prove it based on the definition of $\omega$.
First, the definition of $\omega$ is that it is the smallest inductive set. That is, it is a set I such that $\emptyset \in I$ and for all $x \in I$, $x \cup \{ x \} \in I$, and such that if $J$ satisfies those, then $I \subseteq J$.
First we show that $\omega$ is transitive. Let $T = \{ n \in \omega : n \subseteq \omega \}$. Clearly if $T = \omega$ then $\omega$ is transitive. We show that $T$ is inductive. First $\emptyset \in T$ by definition. Then if $x \in T$, then $x \subseteq \omega$ and $x \in \omega$, therefore $x \cup \{ x \} \subseteq \omega$, so $x \cup \{ x \} \in T$. Hence $T$ is inductive and therefore $T = \omega$.
Now there are a couple of things we can show. Most commonly, we would show that $\in$ is a well order on $\omega$. Another equivalent definition (in the presence of the axiom of foundation) of an ordinal is a transitive set of transitive sets, this may be easier to show. So we let $T = \{ n \in \omega : \forall x \in n \: x \subseteq n \}$. Then again, we wish to show that $T = \omega$, and we do so by showing that $T$ is inductive. $\emptyset \in T$ again vacuously. If $n \in T$, then let $x \in n \cup \{ n \}$. Either $x \in n$, and therefore $x \subseteq n \implies x \subseteq n \cup \{ n \}$, or $x = n$, and therefore $x \subseteq n \cup \{ n \}$. Hence $T$ is inductive so again $T = \omega$, and therefore $\omega$ is a transitive set of transitive sets.