Finite Chain is Order-Isomorphic to Finite Ordinal

Theorem
Let $(S, \preceq)$ be an ordered set.

Let $C$ be a finite chain in $S$.

Then for some finite ordinal $\mathbf n$:
 * $(C, \restriction_C \preceq)$ is order-isomorphic to $\mathbf n$.

That is:
 * $(C, \restriction_C \preceq)$ is order-isomorphic to $\{k \in \N: k < n\}$.

Proof
By the definition of finite set:
 * there is an $n \in \N$ such that there is a bijection $f: C \to \N_n$, where $\N_n = \{k \in \N: k \lt n\}$.

This $n$ is unique.

Define a function $g:\N_n \to C$ recursively by letting
 * $g(0) = \min C$
 * $g(k+1) = \min \left({ C \setminus g(N_k) }\right)$