The following exercise comes from Enderton's Elements of Set Theory, Chapter 7, Exercise 4:
Let $<$ be the usual ordering on the set $P$ of positive integers. For $n$ in $P$, let $f(n)$ be the number of distinct prime factors of $n$. Define the binary relation $R$ by $mRn$ iff either $f(m) < f(n)$ or [$f(m) = f(n)$ and $m < n$]. Show that $R$ is a well ordering on $P$.
I understand what this ordering "looks like," and I was able to prove that it's a total order. However, I'm having trouble proving that $R$ is well-founded. Intuitively, I understand why this is the case: given a set $A$ of positive integers, choose the element $n$ of $A$ with the least number of unique prime factors -- that is, let $n$ be the least element of $f[A]$ (this is possible because $f[A]$ is a nonempty set of natural numbers). Then, choose the least (with respect to the usual ordering on the integers) element $m$ of $f^{-1}(\{n\}) \cap A$ (this is possible because $f^{-1}(\{n\}) \cap A$ is a nonempty set of positive integers). $m$ will be the $R$-least element of $A$, so $A$ is well-founded.
Is this a correct proof? It seems correct from an intuitive point of view, but I also think it's overly complicated. Additionally, I'm concerned it doesn't actually rely on the definition of $f$, so it would seem to apply to any function that maps $P$ to $\omega$.