Divisor Count Function from Prime Decomposition/Proof 1

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

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