Divisor Count Function from Prime Decomposition

Theorem
Let $$n$$ be an integer such that $$n \ge 2$$, with prime decomposition $$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$$.

Let $$\tau \left({n}\right)$$ be the tau function of $$n$$.

Then $$\tau \left({n}\right) = \prod_{j=1}^r \left({k_j + 1}\right)$$.

Proof
We have $$d \backslash n \Longrightarrow \forall i: 1 \le i \le r: d = p_1^{l_1} p_2^{l_2} \ldots p_1^{l_1}, 0 \le l_i \le k_i$$.

For each $$i$$, there are $$k_i+1$$ choices for $$l_i$$, making $$\left({k_1 + 1}\right) \left({k_2 + 1}\right) \cdots \left({k_r + 1}\right)$$ choices in all.

By the Fundamental Theorem of Arithmetic and hence the uniqueness of prime decomposition, each of these choices results in a different number, therefore a distinct divisor.

Alternatively, the result follows immediately from Tau of Power of Prime and Tau Function is Multiplicative‎.