Assume $f(n)$ is multiplicative and that $\sum_1 ^{\infty} f(n)$ converges absolutely. Show that $$\sum_1 ^{\infty} f(n) = \prod_p \sum_{k=0}^{\infty} f(p^k).$$
Hint: Show that if $f(n)$ is also completely multiplicative then $f(p) \neq 1$ for every $p$ and $$\sum_1 ^{\infty} f(n) = \prod _p (1-f(p))^{-1}.$$
Is there a way to show the first equality without using the hint? I also do not know how to show the equality in the hint. Any ideas on how to proceed and conclude? This is in Leveque's Number Theory. The section with this problem has a theorem that states $\prod_p (1-p^{-s})^{-1} = \sum_{k=1}^{\infty} \frac{1}{k^s}$ for $s>1$ and that there are infinitely many primes of the form $4k+1$. Other than these, there is not much else discussed.
Definition: A completely multiplicative function is a function whose domain is $\mathbb{N}$ with $f(1)=1$ and $f(ab)=f(a)f(b)$ for all $a,b \in \mathbb{N}$.
We have $f(p^k) = f(p)^k$ for a completely multiplicative function, thus $$ \prod_p \sum_{k=0}^{\infty} f(p^k) =\prod_p \sum_{k=0}^{\infty} f(p)^k = \prod_p (1 - f(p))^{-1} $$ by the geometric sum formula. Of course one should check that $|f(p)| < 1$, but this follows from the fact that the series converges absolutely, since if this value were larger then $|f(p^k)| = |f(p)|^k \geq 1$ for all $p^k$, but the terms of the sum must go to zero!