Multiplicative functions are determined by their prime-power values

By 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, the prime-power factors $p_i^{e_i}$ are pairwise coprime and their product is $n$. Repeatedly applying multiplicativity gives $f(n)=f({\prod }_{i<r}{p}_i^{e_i})={\prod }_{i<r}f(p_i^{e_i})$. For $n=1$ this is the empty product, so it reads $f(1)=1$.

For the converse, define $F(1):=1$ and, for $n>1$ with canonical factorization $n={\prod }_{i<r}{p}_i^{e_i}$, define $F(n):={\prod }_{i<r}c(p_i^{e_i})$, where $c(p^e)$ is the prescribed prime-power datum. This is well defined because 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 uniquely determines the primes and exponents.

If $\gcd (m,n)=1$, then the canonical factorization of $mn$ is exactly the disjoint union of the canonical factorizations of $m$ and $n$, again by 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. Therefore the defining products for $F(mn)$, $F(m)$, and $F(n)$ split as $F(mn)=F(m)F(n)$, so $F$ is multiplicative.

Any multiplicative function with the prescribed prime-power values must satisfy the formula of step 1.1, so it agrees with $F$ on every positive integer. Thus the extension is unique.