This may be off base, but maybe this is the kind of thing you're looking for.

Start with a positive integer, like 5, and write it entirely in base 2. That is, write every number that appears in base 2, so 5=2²²⁰+2⁰. Now change all the 2's to 3's and subtract 1 to get 27. Write that entirely in base 3, so 27=3³³⁰. Change all the 3's to 4's and subtract 1. Keep going, always replacing n by (n+1), subtracting 1, and writing the number "entirely in base (n+1)".

The result is that no matter what positive integer you start with, you'll eventually get to 0. The usual proof is a complexity argument that uses ω, but maybe you could do it without.

This may be off base, but maybe this is the kind of thing you're looking for.

Start with a positive integer, like 5, and write it entirely in base 2. That is, write every number that appears in base 2, so 5=2²²⁰+2⁰. Now change all the 2's to 3's and subtract 1 to get 27. Write that entirely in base 3, so 27=3³³⁰. Change all the 3's to 4's and subtract 1. Keep going, always replacing n by (n+1), subtracting 1, and writing the number "entirely in base (n+1)".

The result is that no matter what positive integer you start with, you'll eventually get to 0. The usual proof is a complexity argument that uses ω.

Edit: this result is called Goodstein's Theorem