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 $q \ne w$ 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 $n'$ with $n = \pi^{v} n'$, $n' \ne 0$ and $\pi \nmid n'$.

For a prime $w$ and $k \in \mathbb{N}$: $v_w(w^{k}) = k$, and $v_q(w^{k}) = 0$ for every prime $q \ne w$. 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.

$n' \ge 1$: $\pi^{v} \ge 1 > 0$ and $n > 0$, so $n' \le 0$ would make $\pi^{v} n' \le 0 < n$; hence $n' > 0$, and a positive integer is $\ge 1$ because $n' = \iota(j)$ with $j \ne 0$, 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 $M' := \prod_{i<r} p_i^{\,e_i}$ we have $M = M' \cdot p_r^{\,e_r}$, with $M' \ge 1 > 0$ and $p_r^{\,e_r} \ge 1 > 0$, so $M \ge 1 > 0$ and both factors are nonzero; additivity then gives $v_u(M) = v_u(M') + v_u(p_r^{\,e_r})$ for every prime $u$. Taking $u = p_j$ with $j < r$ gives $e_j + 0 = e_j$, since $p_j \ne p_r$ 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 $p \restriction t$ is an injective list of primes containing every prime divisor of $n'$. Indeed if $q$ is prime with $q \mid n'$ then $q \mid n$, since $n = \pi^{v} n'$; so $q = p_i$ for some $i < \sigma(t)$; and $q \ne \pi = p_t$, because $\pi \nmid n'$; hence $i \ne t$, and $i < \sigma(t)$ gives $i \le t$, so $i < t$.

For $i < t$ we have $v_{p_i}(n) = v_{p_i}(\pi^{v}) + v_{p_i}(n') = 0 + v_{p_i}(n') = v_{p_i}(n')$, using additivity on the nonzero factors $\pi^{v}$ and $n'$, and step 2.1 with $p_i \ne \pi$, which holds by injectivity since $i < t$.

By step 1.3 applied to $n'$ and $p \restriction t$: $n' = \prod_{i<t} p_i^{\,v_{p_i}(n')}$.

Therefore $\prod_{i<\sigma(t)} p_i^{\,v_{p_i}(n)} = \bigl(\prod_{i<t} p_i^{\,v_{p_i}(n)}\bigr) \cdot \pi^{\,v_{\pi}(n)} = \bigl(\prod_{i<t} p_i^{\,v_{p_i}(n')}\bigr) \cdot \pi^{v} = n' \pi^{v} = n$, 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.