This may be deduced from a more general result that's both simpler to prove and more insightful, viz. the result follows immediately by this frequently applicable multiplicative form of induction, which shows that for multiplicative sets we need only check the generators (here $1$ and primes).
Lemma $\rm\, \mathbb N$ is the only set of naturals containing $\color{#c00}1$ and $\rm\color{#c00}{all\ primes}$ and $\rm\color{#0a0}{closed\ under\ multiplication}$
Proof $\ $ Suppose $\rm\!\ N\subset \mathbb N\:$ has said properties. $ $ We prove by strong induction every natural $\rm\!\ n\in N.\, $ If $\rm\:n\:$ is $\,\color{#c00}1\,$ or $\color{#c00}{\rm prime}$ then by hypothesis $\rm\:n\in N.\:$ Else $\rm\,n\,$ is composite, $ $ hence $\rm\ n = j k\ $ for $\rm\: j,k < n.\:$ By strong induction the smaller $\rm\ j,k\in N,\:$ $\rm\color{#0a0}{thus}$ $\rm\: n = jk\in N.\ \ \ $ QED
This yields the sought result. Let $\rm\!\ N\!\ $ be the set of naturals that have the form $\rm\,2^{\,j}\!\ n\:$ for odd $\rm\:n\in \mathbb N.\ $ $\, 1\!\ $ and all primes $\rm\,p\,$ are in $\rm\!\ N,\, $ by $\, 2 = 2^{\!\ 1}\!\cdot\! 1\,$ and $\rm\, p = 2^{\!\ 0}\!\cdot\! p\,$ for odd $\rm\,p\,$ (and $1).$ $\rm\,N\!\ $ is closed under multiplication by $\rm\, (2^{\,j}\!\ m)\ (2^{\,k}\!\ n) = 2^{\,j+k}\!\ m\!\ n,\: $ with $\rm\!\ m\!\ n\!\ $ odd by $\rm\!\ m,n\!\ $ odd. $\!\ $ So $\rm\ N = \mathbb N\ $ by Lemma.
Corollary $\ $ Every natural $> 0\,$ is a product of primes (i.e. irreducibles).
Proof $\, $ It is clear for $1$ and primes, and products of primes are closed under multiplication.
Remark $ $ One could deduce the sought result from the prior Corollary. Then it reduces to (inductively) proving that a product of $n$ odds remains odd. But this way is not an instructive example of strong induction, since the strong induction is hidden in the proof of the corollary.