On $\omega$ the ordinal $+$ and $\cdot$ are the Peano operations: $\omega$ is closed under ordinal $+$, $\cdot$ and exponentiation, and for naturals $m,n$ the ordinal $m+n$ and $m\cdot n$ are the natural-number sum and product

Write $+$, $\cdot$ and ${\alpha }^{\beta }$ for the ordinal operations (Ordinal addition $\alpha +\beta$, Ordinal multiplication $\alpha \cdot \beta$, Ordinal exponentiation ${\alpha }^{\beta }$, with the conventions ${\alpha }^0=1$ and $0^0=1$), and ${+}_{\mathbb{N}}$, ${\cdot }_{\mathbb{N}}$ for the natural-number operations defined by Peano recursion (Addition of natural numbers, Multiplication of natural numbers). Let $m,n\in \mathbb{N}=\omega$. Then:

(a) Closure. $m+n$, $m\cdot n$ and $m^n$ all lie in $\omega$.

(b) Agreement for $+$ and $\cdot$. $m+n=m{+}_{\mathbb{N}}n$ and $m\cdot n=m{\cdot }_{\mathbb{N}}n$.

(c) Agreement of the orders. For $m,n\in \omega$, $m\in n$ if and only if $m<n$ in the additive order of Order on the natural numbers. This is claim (i) of $\omega$ is the least limit ordinal and is cited, not reproved.

No agreement is claimed for exponentiation. The dictionary drawn here is with construction-of-the-natural-numbers, which defines addition and multiplication and no exponentiation, and nothing among this page's declared prerequisites supplies a natural-number power for the ordinal power to be compared with. What clause (a) says about $m^n$ is only that the ordinal power of two naturals is again a natural.

This item is the dictionary between the two arithmetics on $\mathbb{N}$. Without it the library would carry two unrelated operations written with the same symbol on the same set. No choice principle is used.

Why the limit clause never fires below $\omega$. Every ordinal in $\omega$ is $0$ or a successor ($\omega$ is the least limit ordinal, claim (iv)), so the two remaining clauses of each ordinal recursion are exactly the two Peano clauses of Addition of natural numbers and Multiplication of natural numbers. That is the whole reason the two arithmetics agree, and it is also the precise sense in which ordinal arithmetic extends rather than replaces the arithmetic of $\mathbb{N}$.

The agreement stops immediately above $\omega$. The natural-number operations are commutative; the ordinal operations are not, and the failure begins at the first infinite ordinal, with $1+\omega =\omega \mathrm{\ne }\omega +1$ (FALSE: ordinal addition is commutative). So this item says the ordinal operations restrict correctly, and says nothing about their behaviour anywhere else.

Exponentiation is closure only. construction-of-the-natural-numbers has no exponentiation, and no prerequisite of this page supplies one, so there is no natural-number power here for the ordinal power to agree with and clause (a) is all that this page claims. Wherever in the library a natural-number exponentiation with the clauses $m^0=1$ and $m^{n^{+}}=m^n\cdot m$ is available, the corresponding agreement is a one-line induction of exactly the shape of step 4.1, on top of claim (b) for the product; it is not carried out here only because this page does not declare the page that mints it as a prerequisite.

What would go wrong without this item. The symbol $+$ would denote two different functions on $\mathbb{N}$, one defined in construction-of-the-natural-numbers and one here, with nothing connecting them. Every later computation mixing finite and infinite ordinals, such as the coefficients of a Cantor normal form (Cantor normal form: every nonzero ordinal is ${\omega }^{{\beta }_0}\cdot c_0+\cdots +{\omega }^{{\beta }_{k-1}}\cdot c_{k-1}$ with ${\beta }_0>\cdots >{\beta }_{k-1}$ and each $c_i$ a nonzero natural number, in exactly one way) or the value $2^{\omega }=\omega$ (FALSE: the ordinal $2^{\omega }$ is uncountable), silently uses the identification proved here.