Value of Multiplicative Function is Product of Values of Prime Power Factors

Theorem
Let $f: \Z \to \Z$ be a multiplicative function.

Let $n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$ be the prime decomposition of $n$.

Then:


 * $f \left({n}\right) = f \left({p_1^{k_1}}\right) f \left({p_2^{k_2}}\right) \cdots f \left({p_r^{k_r}}\right)$

Proof
We have $n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$.

We also have $\forall i, j \in \left[{1 \,. \, . \, n}\right]: i \ne j \implies p_i^{k_i} \perp p_j^{k_j}$.

So $f \left({p_i^{k_i} p_j^{k_j}}\right) = f \left({p_i^{k_i}}\right) f \left({p_j^{k_j}}\right)$.

It is a simple inductive process to show that $f \left({n}\right) = f \left({p_1^{k_1}}\right) f \left({p_2^{k_2}}\right) \cdots f \left({p_r^{k_r}}\right)$.