# Clarification on proof for the product form of a sum on a completely multiplicative function

I am working my way through the theorem on ProofWiki, the theorem states the following:

Let $$f$$ be a completely multiplicative arithmetic function.

Let $$\sum_{n=1}^{\infty} f(n)$$ be absolutely convergent.

Then: $$\sum_{n=1}^\infty f(n) = \prod_p \frac1{1-f(p)}$$ where $$p$$ ranges over the primes.

In the proof of this, they begin by using the complete multiplicative property of $$f$$ to say:

$$f(p^k) = f(p)^k$$ for all primes $$p$$, which implies $$f(p) < 1$$ for all primes $$p$$.

I don't see the implication. Could anyone elaborate this for me? It is obvious that $$f(p) < 1$$ for all but finitely many primes, but this follows from the convergence of the series, not the multiplicative property. What am I missing? Thanks!

Assume to the contrary that there is a prime $p$ with $f(p) \geq 1$. Then $f(p^{k})=f(p)^{k}$, $\forall k \in \mathbb{N}$, a contradiction, since then there are infinitely many values of $f$ greater or equal to $1$.