Sum of divisors in terms of their prime factors

Question

The sum of divisors of a number is given by the following expressions:

$$\sigma(n)= \prod_{i=1}^l\left(1+p_i+{p_i}^2+··· +{p_i}^{a_i}\right)$$ $$\sigma(n)=\prod_{i=1}^l\left(\frac{{p_i}^{a_i+1}-1}{p_i-1}\right)$$

for

$$n=\prod_{i=1}^l{p_i}^{a_i}$$

What I'm having trouble understanding is how the first expression can be turned into the second, and while I'm sure the answer should be quite simple I can't see it. Could someone explain it to me?

Answer 1

Using the formula for a geometric series (initial term $1$, ratio $p_i$),

$1+p_i+{p_i}^2+··· +{p_i}^{a_i} = \dfrac{{p_i}^{a_i+1}-1}{p_i-1}. $

You could verify that

$(p_i-1)(p_i^{a_i}+p_i^{a_{i-1}}\cdots+p_i+1)$

$=p_i^{a_i+1}-p_i^{a_i}+p_i^{a_i}-p_i^{a_{i-1}}+\cdots+p_i^2-p_i+p_i-1$

$=p_i^{a_i+1}-1.$