For $n\ge 1$ and any injective list $p:r\to \mathbb{Z}$ of primes containing every prime divisor of $n$, one has $n={\prod }_{i<r}{p}_i^{v_{p_i}(n)}$; the exponents are determined by $n$, and $v_q(n)=0$ for every prime $q$ outside the list

Distinct primes do not divide one another: if $q$ and $w$ are primes with $q\mid w$, then $q$ is a positive divisor of $w$ (as $q>1>0$) other than $1$, so $q=w$. Consequently $v_q(w)=0$ whenever $q\mathrm{\ne }w$ are primes.

$Q(0)$ holds: an empty list contains no prime, so $n$ has no prime divisor at all; if $n>1$ it would have one, so $n=1$, and ${\prod }_{i<0}{p}_i^{v_{p_i}(n)}=1=n$.

Let $t\in \mathbb{N}$ and assume $Q(t)$.

Let $n\ge 1$ and let $p:\sigma (t)\to \mathbb{Z}$ be an injective list of primes containing every prime divisor of $n$. Write $\pi :=p_t$, $v:=v_{\pi }(n)$, and fix $n^{\mathrm{\prime }}$ with $n={\pi }^v{n}^{\mathrm{\prime }}$, $n^{\mathrm{\prime }}\mathrm{\ne }0$ and $\pi \nmid n^{\mathrm{\prime }}$.

For a prime $w$ and $k\in \mathbb{N}$: $v_w(w^k)=k$, and $v_q(w^k)=0$ for every prime $q\mathrm{\ne }w$. Both sets of $k$ for which these hold contain $0$, since $w^0=1$ and $v_u(1)=0$ for every prime $u$; and if they hold at $k$ then, $w^k$ and $w$ being nonzero, additivity gives $v_w(w^{\sigma (k)})=v_w(w^{k}w)=v_w(w^k)+v_w(w)=k+1=\sigma (k)$ and $v_q(w^{\sigma (k)})=v_q(w^k)+v_q(w)=0+0=0$. Induction finishes both.

$n^{\mathrm{\prime }}\ge 1$: ${\pi }^v\ge 1>0$ and $n>0$, so $n^{\mathrm{\prime }}\le 0$ would make ${\pi }^v{n}^{\mathrm{\prime }}\le 0<n$; hence $n^{\mathrm{\prime }}>0$, and a positive integer is $\ge 1$ because $n^{\mathrm{\prime }}=\iota (j)$ with $j\mathrm{\ne }0$, so $1=\sigma (0)\le j$.

Clause 3. Let $p:r\to \mathbb{Z}$ be an injective list of primes, $e:r\to \mathbb{N}$, and $M:={\prod }_{i<r}{p}_i^{e_i}$; we claim $M\ge 1$, that $v_{p_j}(M)=e_j$ for every $j<r$, and that $v_q(M)=0$ for every prime $q$ off the list. Let $T$ be the set of $r\in \mathbb{N}$ for which this holds for all such $p$ and $e$. Then $0\in T$: the empty product is $1$, which is $\ge 1$, has $v_q(1)=0$ for every prime $q$, and imposes no condition on indices. Suppose $r\in T$ and let $p:\sigma (r)\to \mathbb{Z}$ be injective with primes and $e:\sigma (r)\to \mathbb{N}$. Writing $M^{\mathrm{\prime }}:={\prod }_{i<r}{p}_i^{e_i}$ we have $M=M^{\mathrm{\prime }}\cdot p_r^{e_r}$, with $M^{\mathrm{\prime }}\ge 1>0$ and $p_r^{e_r}\ge 1>0$, so $M\ge 1>0$ and both factors are nonzero; additivity then gives $v_u(M)=v_u(M^{\mathrm{\prime }})+v_u(p_r^{e_r})$ for every prime $u$. Taking $u=p_j$ with $j<r$ gives $e_j+0=e_j$, since $p_j\mathrm{\ne }{p}_r$ by injectivity; taking $u=p_r$ gives $0+e_r=e_r$, since $p_r$ is off the list $p_0,\dots ,p_{r-1}$; and taking a prime $u$ off the whole list gives $0+0=0$. So $\sigma (r)\in T$, and $T=\mathbb{N}$ by induction.

The restriction $p\upharpoonright t$ is an injective list of primes containing every prime divisor of $n^{\mathrm{\prime }}$. Indeed if $q$ is prime with $q\mid n^{\mathrm{\prime }}$ then $q\mid n$, since $n={\pi }^v{n}^{\mathrm{\prime }}$; so $q=p_i$ for some $i<\sigma (t)$; and $q\mathrm{\ne }\pi =p_t$, because $\pi \nmid n^{\mathrm{\prime }}$; hence $i\mathrm{\ne }t$, and $i<\sigma (t)$ gives $i\le t$, so $i<t$.

For $i<t$ we have $v_{p_i}(n)=v_{p_i}({\pi }^v)+v_{p_i}(n^{\mathrm{\prime }})=0+v_{p_i}(n^{\mathrm{\prime }})=v_{p_i}(n^{\mathrm{\prime }})$, using additivity on the nonzero factors ${\pi }^v$ and $n^{\mathrm{\prime }}$, and step 2.1 with $p_i\mathrm{\ne }\pi$, which holds by injectivity since $i<t$.

By step 1.3 applied to $n^{\mathrm{\prime }}$ and $p\upharpoonright t$: $n^{\mathrm{\prime }}={\prod }_{i<t}{p}_i^{v_{p_i}(n^{\mathrm{\prime }})}$.

Therefore ${\prod }_{i<\sigma (t)}{p}_i^{v_{p_i}(n)}=\left({\prod }_{i<t}{p}_i^{v_{p_i}(n)}\right)\cdot {\pi }^{v_{\pi }(n)}=\left({\prod }_{i<t}{p}_i^{v_{p_i}(n^{\mathrm{\prime }})}\right)\cdot {\pi }^v=n^{\mathrm{\prime }}{\pi }^v=n$, so $Q(\sigma (t))$ holds.

With step 1.2 as base, induction gives $Q(r)$ for every $r\in \mathbb{N}$, which is clause 1. Clause 2 follows because a prime $q$ off the list does not divide $n$ — otherwise it would be a prime divisor of $n$ and hence on the list — so $v_q(n)=0$; and clause 3 is step 3.1.