$$\color{blue}{\text{PROBLEM}}$$
Show every natural number $n\geq 2$ has a prime factorization.
$$\text{TYPICAL SOLUTION}$$
Base case: $2$ is prime, so it is its own prime factorization.
Inductive step: Suppose for all $k\geq 2$ and $k<n$, then $k$ has a prime factorization. Either $n$ is prime or not prime.
$\hspace{3cm}$ If prime:
$\hspace{4cm}$$n$ is its own prime factorization.
$\hspace{3cm}$ If not:
$\hspace{4cm}$$n=ab$ where $a,b>1$ and $a,b<n$,
so by the hypothesis of $\color{red}{\text{strong induction}}$$^{\dagger}$ $a,b$ both have a prime factorization, so their product is a prime factorization of $n$.
$$\text{MY QUESTION}$$
How can I prove this using the fact that every non-empty subset of $\mathbb{N}_0$ has a least element? Here, $\mathbb{N}_0$ is the set $\mathbb{N}\cup 0$. More specifically though, how can we use my definition for this (see below) to prove the $\color{blue}{\text{PROBLEM}}$?
$$(\forall {\scr{S}}\subseteq \mathbb{N}_0)[({\scr{S}}\neq \emptyset)\implies (\exists s\in{\scr{S}})[(\forall n\in{\scr{S}})(s \leq n)]]$$
$$\text{NOTES}$$
$^{\dagger}$$\color{red}{Strong\text{ }induction}$:
$$\forall {\scr{P}}(n)[(\forall n\in \mathbb{N}_0)[(\forall k \in \mathbb{N}_{<n})(k\in {\scr{P}}(n))]\implies (n\in {\scr{P}}(n)) \implies (\forall n\in\mathbb{N}_0)(n\in {\scr{P}}(n))]$$
$\tiny{{\scr{P}}(n):=\text{"$n$ is in the set ${\scr{P}}$}}$
Suppose the claim is false and let $\;\emptyset\neq H\subset\Bbb N\setminus\{0,1\}\;$ be the set of all elements which have no prime factorization.
Let $\;h\in H\;$ be the first element in $\;H\;$ ; this element cannot be prime as then it'd be its own prime factorization, contradicting the fact that $\;h\in H\;$ . Thus, there exist $\;2\le a,b<h\;$ s.t. $\;ab=h\;$.
But $\;a,b<h\implies a,b\notin H\implies a,b\;$ have a prime factorization, and thus does $\;h\;$ ...