Divisor Sum of Square-Free Integer

Theorem
Let $n$ be an integer such that $n \ge 2$.

Let $n$ be square-free.

Let the prime decomposition of $n$ be:
 * $\ds n = \prod_{1 \mathop \le i \mathop \le r} p_i = p_1 p_2 \cdots p_r$

Let $\map {\sigma_1} n$ be the divisor sum of $n$.

That is, let $\map {\sigma_1} n$ be the sum of all positive divisors of $n$.

Then:
 * $\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} p_i + 1$