Divisor Sum of Square-Free Integer/Proof 2

Proof
From Divisor Sum of Integer:
 * $\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$

where each of the $k_i$s are equal to $1$;

Hence:
 * $\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^2 - 1} {p_i - 1}$

But from Difference of Two Squares:
 * $\dfrac {p_i^2 - 1} {p_i - 1} = \dfrac {\paren {p_i + 1} \paren {p_i i 1} } {p_i - 1} = p_i + 1$

Hence the result.