Justification of restricting to functions with finite support in the definition of ordinal exponentiation

Question

For ordinals $\alpha$ and $\beta$, the ordinal $\beta^\alpha$ is defined to be the epsilon-image of the well-ordered structure $\langle F,<\rangle$, where $F$ consists of the functions $f:\alpha\rightarrow\beta$ with finite support and $<$ is the antilexicographical order. I understand that dropping the constraint of having finite support destroys the well-orderedness of $F$. My question is about the justification of the choice of this constraint. For instance, how do we know that this particular choice of contraint faithfully generalizes the idea/intuition of the "length" of $\alpha\times \beta$ that is well ordered by the antilexicographical order in the definition ordinal multiplication? Can it be the case that the constraint of finite support deletes too many functions from $F$ so that the resulting well-ordered set becomes shorter than what it is supposed to be (i.e., without the constraint)? I hope my question makes sense. Please help. Thanks.

EDIT: I appreciate Alessandro's answer, but I am looking for a more direct justification. As a matter of fact, even the idea of imposing any constraint on the functions in $F$ troubles me, since the resulting $F$ is different from what is generalized from the definition of ordinal multiplication.

Answer 1

A different way to define $\alpha^\beta$ is by transfinite recursion on $\beta$ as follows:

• If $\beta=0$, $\alpha^\beta=1$.
• If $\beta=\gamma+1$ for some $\gamma$, $\alpha^\beta=\alpha^\gamma\cdot\beta$
• If $\beta$ is a limit ordinal, $\alpha^\beta=\sup_{\gamma<\beta}\alpha^\gamma$

The first case is clear, the second generalizes the idea that exponentiation is repeated multiplication, while the third says that exponentiation is continuous. If you agree that this definition makes sense intuitively then using finite support functions is the correct choice, because it agrees with this definition.