Why do we mention that well-ordering principle implies there is a least element? It seems to me that the purpose of the well-ordering principle is to prove that something that is a natural number exists and proving that something exists really has nothing to do with having a least element.. unless of course we must specify what that least element is... or is bounded within a range.  We first prove that a set that is a subset of the naturals is non-empty by naming an example of when a condition of that set is satisfied. show that a natural number q satisfies property P(q),  Then we say that because there is a set with a least element, we don't care about how many elements exactly there are in the set of whether there actually is a least element or what that least element is... therefore there exists some natural number satisfying some property. 
I feel like we never use the well-ordering principle to prove that there is a least element, which is kind of like a lower-bound, an infimum or to address what the lower bound is, or to show how many elements there are in a set, we just try to show  if there is non-empty set then there exists a q $\in$ 
$\mathbb{N}$. Why do we even mention that the well-ordering principle shows that there is a least element if the most frequent use of the WOP is to show that something exists? If we use the WOP as a tool to show existence, lets say in euclidean division algorithm n=ad+c, there exists a c, why do we care whether they is a least element? I think learning the fact that WOP shows that there exists some least element is useless. We should just show non-empty is the subset of natural numbers and therefore there exists a remainder from  Euclidean division.
I feel like the least element implication of the WOP is a useless fact to know and it's not important to write in a proof if the only practical purpose of WOP is to show existence. 
 A: I don't understand what you're asking, so let me prove a classical theorem using the well ordering principle, and you can tell me if there is some kind of problem with the way I'm using it.
Let $n$ be a natural number which is $\geq 2$.  We say that $n$ is composite if there exist natural numbers $a$ and $b$, both less than $n$, such that $n = ab$.  Otherwise, $n$ is called prime.
Theorem: Every natural number $n \geq 2$ can be written as a product of prime numbers.  That is, there exists a natural number $k \geq 1$, and prime numbers $p_1, ... , p_k$, such that $n = p_1 \cdots p_k$.  
Proof: Suppose not.  Then the set $S$ of natural numbers which are $\geq 2$ and which cannot be written as a product of prime numbers is not empty.  Being a nonempty subset of the natural numbers, the well ordering principle tells us that $S$ must contain a least element $s$.  
Now $s$ is itself not prime, so there exist natural numbers $a$ and $b$, both less than $s$, such that $s = ab$.  It is clear that $a$ and $b$ are both $\geq 2$.  Since they are both less than $s$, they do not lie in $S$, and so they are both expressible as products of prime numbers.  Hence $s$ is a product of prime numbers, contradiction.  $\blacksquare$
