Divisor Count Function is Multiplicative

Theorem
The divisor counting function:
 * $\ds \tau: \Z_{>0} \to \Z_{>0}: \map \tau n = \sum_{d \mathop \divides n} 1$

is multiplicative.

Proof
Let $f_1: \Z_{>0} \to \Z_{>0}$ be the constant function:
 * $\forall n \in \Z_{>0}: \map {f_1} n = 1$

Thus we have:
 * $\ds \map \tau n = \sum_{d \mathop \divides n} 1 = \sum_{d \mathop \divides n} \map {f_1} d$

But from Unity Function is Completely Multiplicative, $f_1$ is multiplicative.

The result follows from Sum Over Divisors of Multiplicative Function.