TITLE: Has the exponentiation of ordinals a nice geometric model?
QUESTION [16 upvotes]: It is well known that the sum $\alpha+\beta$ of two ordinals $\alpha,\beta$ can be defined "geometrically" as the order type of the sum $(\{0\}\times \alpha)\cup(\{1\}\times\beta)$ endowed with the lexicographic order.
Also the product $\alpha\cdot\beta$ of ordinals $\alpha,\beta$ is the order type of the Cartesian product $\beta\times\alpha$ endowed with the lexicographic order.

What about the exponentiation of ordinals?
Does $\alpha^\beta$ have some nice "geometric'' or combinatorial model?
Maybe as some set of (partial) functions endowed with a suitable well-order?

REPLY [21 votes]: Ordinal exponentiation is a special case of linear order exponentiation. For any linear order $L$, element $a\in L$, and ordinal $\beta$ we can define the $\beta$th power of $L$ at $a$, which I'll call "$L_a^\beta$," as the set of functions $f:\beta\rightarrow L$ such that all but finitely many $\alpha\in\beta$ have $f(\alpha)=a$. The ordering on this set is given by looking at the last point of difference: $$f\trianglelefteq g\iff f=g\mbox{ or } f(\max\{x:f(x)\not=g(x)\})<g(\max\{x:f(x)\not=g(x)\}).$$
For ordinals $\alpha,\beta$ we have $\alpha^\beta=\alpha_0^\beta$. Rosenstein's book treats this in some detail (and is generally an awesome book all-around - it's a huge tragedy that it's so hard to find).
What I like about this definition is that it very nicely complements the definition of cardinal exponentiation: they start with the same basic idea of counting functions between sets, but ordinal exponentiation is the "finite support" version.