Divisor Count Function is Multiplicative

Theorem
The tau function:
 * $$\tau: \Z^*_+ \to \Z^*_+: \tau \left({n}\right) = \sum_{d \backslash n} 1$$

is multiplicative.

Proof
Let $$f_1: \Z^*_+ \to \Z^*_+$$ be the constant function:
 * $$\forall n \in \Z^*_+: f_1 \left({n}\right) = 1$$.

Thus we have:
 * $$\tau \left({n}\right) = \sum_{d \backslash n} 1 = \sum_{d \backslash n} f_1 \left({d}\right)$$.

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

The result follows from Sum Over Divisors of Multiplicative Function.