I wish to prove that the sum of the divisors function, $\sigma(x)$, is multiplicative, i.e. $$\sigma(m\cdot n)=\sigma(m)\cdot\sigma(n)$$ if $gcd(m,n)=1$.
I start by claiming that since $m$ and $n$ can be written uniquely as a product of prime factors, let
$$m=\prod_{i=1}^k \, p_i = p_1 \cdot p_2 \cdots p_k$$
$$n=\prod_{j=1}^l \, q_j = q_1 \cdot q_2 \cdots q_l$$
such that $p_i \neq q_j$ for any $i,j \in \mathbb{N}$. Then the sum of the divisors of $m$ is
\begin{align} \sigma(m) = \,\,& p_1 +p_1 \cdot p_2 + p_1 \cdot p_2 \cdot p_3 + \cdots + m + \\ &p_2+p_2 \cdot p_3 + p_2 \cdot p_3 \cdot p_4 + \cdots + \prod_{i=2}^k \,p_i + \\ &p_3+p_3 \cdot p_4 + p_3 \cdot p_4 \cdot p_5 + \cdots + \prod_{i=3}^k \,p_i + \\ &\vdots \\ &+ p_k +1 \end{align}
Or, $$\sigma(m)=\sum_{z=1}^k \left(\sum_{h=z}^{k}\left(\prod_{i=z}^h \,p_i\right)\right)+1.$$
Similarly,
$$\sigma(n)=\sum_{y=1}^l \left(\sum_{r=y}^{l}\left(\prod_{j=y}^r \,q_j\right)\right)+1.$$
Then the product $m \cdot n$ will have for divisors every divisor of $m$ excluding $1$, every divisor of $n$ excluding $1$, $1$, and every possible combination of $m$'s divisors with $n$'s divisors. The combinations are given by the product of the two expressions for $\sigma(m)$ and $\sigma(n)$ less one. Then,
\begin{align} \sigma(m \cdot n)=&\sum_{z=1}^k \left(\sum_{h=z}^{k}\left(\prod_{i=z}^h \,p_i\right)\right) \cdot \sum_{y=1}^l \left(\sum_{r=y}^{l}\left(\prod_{j=y}^r \,q_j\right)\right) + \\[2ex] &\sum_{z=1}^k \left(\sum_{h=z}^{k}\left(\prod_{i=z}^h \,p_i\right)\right) + \sum_{y=1}^l \left(\sum_{r=y}^{l}\left(\prod_{j=y}^r \,q_j\right)\right) + 1 \end{align}
Thus,
\begin{align} \sigma(m \cdot n) &= \left[\sum_{z=1}^k \left(\sum_{h=z}^{k}\left(\prod_{i=z}^h \,p_i\right)\right)+1\right] \cdot \left[\sum_{y=1}^l \left(\sum_{r=y}^{l}\left(\prod_{j=y}^r \,q_j\right)\right)+1 \right] \\[2ex] \sigma(m \cdot n) &=\sigma(m) \cdot \sigma(n) \end{align} $$\tag*{$\blacksquare$}$$ My question is two-fold.
$(a)$ Is this a valid proof? $(b)$ Is the Sigma-Sigma-Pi notation for $\sigma(m)$, $\sigma(n)$ an appropriate notation, i.e. is there a more concise way to express them?