Double counting the number of proper divisors Suppose $n$ is a composite natural number.  Then $n$ has unique prime factorization.  To count the number of proper divisors, simply take the product of the exponents +1 in the prime factorization.
$$n = \prod_{i = 1}^{n} p_i^{a_i}$$
$$\mbox{proper divisors} = \prod_{i = 1}^{n}(a_i+1)$$
Is 1 counted multiple times by doing this?  
For instance, I can choose from $a_0+1$ factors contributed from $p_0$, namely
$ 1, p_0, p_0^2, \dots , p_0^{a_0} $
Don't I count 1 multiple times?

In order to count one, the exponent on each prime factor has to be zero.  So you are only counting it once.
