The sigma function (sum of divisors) multiplicative proof I am trying to prove that $\sigma(p_1^a\cdot p_2^b) =\sigma(p_1^a)\cdot\sigma(p_2^b)$ where $p_1$ and $p_2$ are prime numbers.
We know that $\sigma(p_1^a) = \frac{p_1^{a+1}-1}{p_1-1}$ and $\sigma(p_2^b) = \frac{p_2^{b+1}-1}{p_2-1}$.
Now I am trying to find the divisors of $p_1^a\cdot p_2^b$ and add them:
I found that the divisors are 
$1$, $p_1$, $p_1^{2},\dotsc,p_1^{a}$, $p_2$, $p_2^{2},\dotsc,p_2^{b}$, $p_1\cdot p_2$, $p_1\cdot p_2^2,\dotsc,p_1\cdot p_2^{b},\dotsc,p_1^{a}\cdot p_2,\dotsc,p_1^{a}\cdot p_2^{b}$.                             
Now when we do their summation we get $\sum_{k=0}^a$ $p_1^k$ + $\sum_{k=1}^b$ $p_2^k$ + ($\sum_{k=1}^{a}$ $p_1^k\cdot\sum_{k=1}^b$ $p_2^k$), is this right? If yes I can't reach $\frac{p_1^{a+1}-1}{p_1-1}\cdot\frac{p_2^{b+1}-1}{p_2-1}$.
 A: You are very close.  Just notice that
$$\sum_{k=0}^a p_1^k + \sum_{k=1}^b p_2^k + \left(\sum_{k=1}^{a} p_1^k\cdot\sum_{k=1}^b p_2^k \right) $$
equals
$$
1+\sum_{k=1}^a p_1^k + \sum_{k=1}^b p_2^k + \left(\sum_{k=1}^{a} p_1^k\cdot\sum_{k=1}^b p_2^k \right)=\sum_{k=0}^a p_1^k \cdot \sum_{k=0}^b p_2^k
$$
which is 
$$
\sigma(p_1^a) \sigma(p_2^b)
$$
so you are done.
A: In general if the prime factorization of $n$ is
$$n=p_1^{a_1}p_2^{a_2}\dotsm p_m^{a_m},$$
then the divisors of $n$ will be of the form
$$p_1^{b_1}p_2^{b_2}\dotsm p_m^{b_m}\quad (0\leqslant
b_i\leqslant a_i).$$
Now consider the product
$$(1+p_1+p_1^2+\dotsb+p_1^{a_1})(1+p_2+p_2^2+\dotsb+p_2^{a_2})
\dotsm(1+p_m+p_m^2+\dotsb+p_m^{a_m})$$
which when expanded will contain all the divisors of $n$ in a sum of
the divisors of $n$. As each bracketed term has $(a_i+1)$ terms, these being the values $b_i$ can take, namely $0$ to $a_i$, this gives the
number of divisors of $n$,  $\sigma_0(n)$, alternatively written $s(n)$,  by the product
$$\sigma_0(n)=\prod_{i=1}^m(a_i+1)=(a_1+1)(a_2+1)\dotsm(a_m+1).$$
The sum of the divisors of $n$, $\sigma(n)$ is then
$$
\sigma(n)=\prod_{i=1}^m(1+p_i+p_i^2+\dotsb+p_i^{a_i})
=\prod_{i=1}^m\frac{p_i^{a_i+1}-1}{p_i-1}=\prod_{i=1}^m\sigma(p_i^{a_i}).
$$
A: $$\sigma(n) = \sum_{d \mid n} d$$
If $\gcd(n,m) = 1$ then there is bijection $(d,d') \to dd'$ between the couples of divisors $ d \mid n, d' \mid m$ and the divisors of $ nm$, and hence $$\sigma(n)\sigma(m) = (\sum_{d \mid n} d)(\sum_{d' \mid m} d') = \sum_{d \mid n, \ d' \mid m} dd' = \sum_{k \mid nm} k=\sigma(nm)$$
