Prove Proposition (4.37).
Proposition (4.37). Let the prime decomposition of a natural number $n$ be given by
$$ n = p_1^{k_1} \times p_2^{k_2} \times p_3^{k_3} \times \ldots \times p_m^{k_m} $$
where $p_j$’s are distinct primes. Then
$$ \begin{align} \sigma(n) & = \sigma(p_1^{k_1} \times p_2^{k_2} \times p_3^{k_3} \times \ldots \times p_m^{k_m}) \\ \\ & = \sigma(p_1^{k_1}) \times \sigma(p_2^{k_2}) \times \sigma(p_3^{k_3}) \times \ldots \times \sigma(p_m^{k_m}) \end{align} $$
We start by noting that the divisors of $n$ are of the form
$$ p_1^{a_1} \times p_2^{a_2} \times p_3^{a_3} \times \ldots \times p_m^{a_m} $$
where $0 \le a_i \le k_i$ where $1 \le i \le m$.
The sum of all such divisors is given by the following product. To explain this further, multiplying out the brackets results in summands that represent every possible divisor of the above form.
$$ \begin{align} \sigma(n) & = (\sum_{i=0}^{k_1} p_1^i) \times (\sum_{i=0}^{k_2} p_2^i) \times (\sum_{i=0}^{k_3} p_3^i) \times \ldots \times (\sum_{i=0}^{k_m} p_m^i) \\ \\ & = \sigma(p_1^{k_1}) \times \sigma(p_2^{k_2}) \times \sigma(p_3^{k_3}) \times \ldots \times \sigma(p_m^{k_m}) \end{align}$$
This is the desired result.
Note: the author's solution incorrectly applies Proposition (4.36).