Prime factors + number of Divisors

If you write $n= \prod\limits_{i=1}^n {p_i}^{e_i}$, then the number of divisors of $n$ is the number of different products you can make of the form $\prod\limits_{i=1}^n {p_i}^{f_i}$ where $f_i \leq e_i$. For each prime, there are $e_i+1$ choices for $f_i$ and so the total number of choices is $\prod\limits_{i=1}^n (e_i+1)$ as you suggest.

Let $d(n)$ be the number of divisors for the natural number, $n$.

We begin by writing the number as a product of prime factors: $n = p^aq^br^c...$ then the number of divisors, $d(n) = (a+1)(b+1)(c+1)...$

To prove this, we first consider numbers of the form, $n = p^a$. The divisors are $1, p, p^2, ..., p^a$; that is, $d(p^a)=a+1$.

Now consider $n = p^aq^b$. The divisors would be:

$1, p, p^2, ..., p^a,$

$q, pq, p^2q, ..., p^aq,$

$q^2, pq^2, p^2q^2, ..., p^aq^2,$

$..., ..., ..., ..., ...,$

$q^b, pq^b, p^2q^b, ..., p^aq^b $

Hence we prove that the function, $d(n)$, is multiplicative, and in this particular case, $d(p^aq^b)=(a+1)(b+1)$. It should be clear how this can be extended for any natural number which is written as a product of prime factors.

http://mathschallenge.net/library/number/number_of_divisors