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:
 * $\displaystyle n = \prod_{1 \mathop \le i \mathop \le r} p_i = p_1 p_2 \cdots p_r$

Let $\sigma \left({n}\right)$ be the sigma function of $n$.

That is, let $\sigma \left({n}\right)$ be the sum of all positive divisors of $n$.

Then:
 * $\displaystyle \sigma \left({n}\right) = \prod_{1 \mathop \le i \mathop \le r} p_i + 1$