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 $\frac{p_1^{a+1}-1}{p_1-1}\cdot\frac{p_2^{b+1}-1}{p_2-1}$.

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.
