It almost certainly doesn't, since it's an old and definitely a well-tested theorem, but I don't understand why.
Begging the question, or circular reasoning, as defined in my textbook (of sorts), is:
...the author of a proof uses in his argument the fact that he is supposed to prove.
The well-ordering principle, on the other hand, is defined as:
If S is a nonempty subset of N then there is m ∈ S such that m ≤ x for all x ∈ S. That is, S has a smallest or least element.
Then, as a proof for the well-ordering principle, they write:
We will use contraposition to prove the theorem. That is, by assuming that S has no smallest element we will prove that S = ∅. We will prove that n ∉ S for all n ∈ N. We do this by induction on n. Since S has no smallest element, we have 1 ∉ S. Assume that we have proved that 1, 2, · · · , n ∉ S. We will show that n + 1 ∉ S. If n + 1 ∈ S then n + 1 would be the smallest element of S since 1, 2, 3, · · · , n ∉ S, and this contradicts the assumption that S has no smallest element. Thus, we must have n + 1 ∉ S. Hence, by the principle of mathematical induction, n ∉ S for all n ∈ N. But this leads to S = ∅. This establishes a proof of the theorem
The well-ordering principle is trying to prove that every non-empty set has a smallest element, so why does its proof assume that n + 1, the only element in the set S, would be the smallest one?
I do get that n + 1 would be the smallest element if 1, 2, 3 . . . n do not belong to the set S (obviously), but does that assumption not make use of the well-ordering principle, which it's trying to prove?
(I'm a sophomore undergrad, so, if possible, relatively simple explanations would be appreciated.)