Divisor Counting Function from Prime Decomposition/Proof 1

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

For each $i$, there are $k_i + 1$ choices for $l_i$, making $\paren {k_1 + 1} \paren {k_2 + 1} \cdots \paren {k_r + 1}$ 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.