Declare a well ordered set S bad if it is not isomorphic to an ordinal, and declare it "very bad" if every proper initial segment of S is good.
If S is bad, consider T, a subset of S, defined as follows. For x in S, take the initial segment below x, and if it is bad, include it in T. If T is empty then S is very bad. If T is nonempty then choose x minimal. Replace S with the elements below x, and S is now very bad. Note that S is not the empty set, else it is already an ordinal.
At this point S is very bad, hence each proper initial segment of S has a unique isomorphism h mapping it onto the members of an ordinal. Let a new function j map each x to the ordinal that the elements below x map onto, via h. Thus j maps the members of S onto various ordinals. The ordinal is unique, hence j is a function.
Verify that j respects order. Restrict h to a shorter segment and find the unique isomorphism that maps that segment onto a smaller ordinal. Thus as x decreases in S, j(x) descends down the ordinals.
Let r be the range of j, which is a set by replacement. If q is a subset of r, map it back to S and take the least element. Thus r is well ordered. If y ∈ r and z ∈ y, map the elements of y back to S, restrict h to those elements of S that map to the elements of z, and build a new initial segment in S. The upper bound of this segment maps to z, hence z is in the image of j, and in r. Now r is transitive, and is an ordinal. Every well ordered set is isomorphic to a unique ordinal.
The relation "isomorphic to" is an equivalence relation, defining equivalence classes of well ordered sets. Each equivalence class is represented by an ordinal. If you can well order the members of a set, that ordered set is represented by a specific ordinal. Of course a different well ordering might lead to a different ordinal, but that's another story.