How to define an isomorphism between $^\omega\omega$ and $\omega^\omega$?

Question

Let $^\omega\omega$ be the set of all functions $x: \omega \to \omega$. Define $A = \{x \in ^\omega\omega \; | \; x \text{ has finite support}\}$, where by "finite support" I mean that the set $\{x(n) \; | \; x(n) \not= 0\}$ is finite. Finally, define $\prec_A$ as $x \prec_A y \iff x(n) < y(n)$ for the greatest $n \in \omega$ on which $x, y$ differ (it's not difficult to see that such an $n$ exists). It's rather easy to see that $\langle A, \prec_A \rangle$ is a well-order, whence, by Mostowski Collpase, it follows that it is isomorphic to an ordinal $\gamma$. My problem now is: how do I show that $\gamma = \omega^\omega$? I tried to define directly an isomorphism between $\langle A, \prec_A \rangle$ and $\omega^\omega$, but couldn't get very far. Any ideas?

Answer 1

HINT: Show that for every $n\in\omega$, $\omega^n$ embeds into $A$, and that $A$ itself embeds into $\omega^\omega$.

Answer 2

Some notes: First show that $^n \omega$ is isomorphic to $\omega^n$, the show that all elements of $^n\omega$ are larger that elements of $^m\omega$ for $n >m$.