Jonathan wrote:

> If we do this again, what do we get?

I'm not sure what "do this again" means. I would love to get [hyperoperations](https://en.wikipedia.org/wiki/Hyperoperation#Examples) this way. But there are some puzzling features.

Note that addition and multiplication are associative but exponentiation is not. Composition of morphisms is automatically associative. So, \\( (\mathbb{N},+,0)\\) and \\( (\mathbb{N},\cdot,1)\\) can be regarded as one-object categories but \\( (\mathbb{N},\uparrow,???)\\) cannot, where \\(a \uparrow b = a^b\\) is exponentation.

Exponentiation also lacks an identity, which is why I write \\(???\\): we have \\(a\uparrow 1 = a\\) but not \\(1 \uparrow a = a\\).

For these reasons, we were able to get nice category-theoretic descriptions of
\\( (\mathbb{N},+,0)\\) and \\( (\mathbb{N},\cdot,1)\\), but the pattern must change a bit when we hit exponentation and higher hyperoperations:

* \\( (\mathbb{N},+,0)\\) is isomorphic to \\(\mathbf{N}\\), the free category on the graph

* \\( (\mathbb{N},\cdot,1)\\) is isomorphic to a category I'll call \\(\mathbf{M}\\) with \\(\mathbf{N}\\) as its only object and functors from \\(\mathbf{N}\\) to itself as morphisms.

We could try to charge ahead naively:

**Puzzle.** What is the category with \\(\mathbf{M}\\) as its only object and functors from \\(\mathbf{M}\\) to itself as morphisms?

This could be interesting, and we could keep repeating this. But you are wisely taking a detour: to get exponentation, you're looking at functors from \\(\mathbf{N}\\) to \\(\mathbf{M}\\). What does it mean to "do this again"?