I don’t see anything wrong with the proof, particularly the formula
$\displaystyle \sigma(n)\ =\ \prod_{i\,=\,1}^r\frac{p_i^{k_i+1}-1}{p_i-1}$
Shall we prove this formula? Then perhaps you’ll feel less afraid of using it.
First, we prove that if $\displaystyle \gcd(m,n)=1$ then $\displaystyle \sigma(mn)=\sigma(m)\sigma(n).$ Note that a divisor $\displaystyle d$ of $\displaystyle mn$ can be written as $\displaystyle d=ef$ where $\displaystyle e\mid m$ and $\displaystyle f\mid n$ and $\displaystyle \gcd(e,f)=1.$ Hence
$\displaystyle \sigma(mn)\,=\,\sum_{d\,\mid\,mn}d$
$\displaystyle =\,\sum_{e\,\mid\,m\,f\,\mid\,n}ef$
$\displaystyle =\,\sum_{e\,\mid\,m}e\,\sum_{f\,\mid\,n}f$
$\displaystyle =\,\sigma(m)\sigma(n)$
Next, not that if $\displaystyle p$ is a prime and $\displaystyle k$ a positive integer, $\displaystyle \sigma(p^k)=1+p+p^2+\cdots+p^k=\frac{p^{k+1}-1}{p-1}.$
Combining the two results established above gives the required formula for the sum-of-factors function $\displaystyle \sigma.$