A question about the property of completely multiplicative functions

Question

I am self-studying number theory. In Apostol's number theory textbook, Theorem 2.17 states:

Let $f$ be multiplicative. Then $f$ is completely multiplicative if and only if $$f^{-1}(n)=\mu(n) f(n) \text{ for all } n \geq 1.$$

There is one part of the proof that I do not quite understand so I was hoping to clarify it. The proof says to prove that $f$ is completely multiplicative, it suffices to prove $f(p^a)=f(p)^a$. It also says $$\sum_{d |n} \mu(d)f(d)f(\frac{n}{d})=0$$ I understand these parts. However, later it writes $n=p^a$ and concludes:

$$\mu(1)f(1)f(p^a) + \mu(p)f(p)f(p^{a-1}) = 0 \quad (*)$$

from which we find $f(p^a) = f(p)f(p^{a-1})$. This implies $f(p^a) = f(p)^a$.

Here are two questions that I have:

1. Why are there only two terms in the summation $(*)$? Because we do not know the value of $a$, shouldn't the intermediate divisors of $p^a$ be included in $(*)$? Since the divisors of $p^a$ are $\{1,p,p^2,\dots,p^a\}$.

2. How does $f(p^a) = f(p)f(p^{a-1})$ imply $f(p^a) = f(p)^a$? I understand where the $f(p^a) = f(p)f(p^{a-1})$ comes from, but I do not see how it implies $f(p^a) = f(p)^a$, since at this stage we still do know if $f$ is completely multiplicative and $\gcd(p, p^{a-1})\neq 1.$ I think I am missing something trivial.

Answer 1

1. Because in the sum $$ 0 = \sum_{i=0}^a \mu(p^i)f(p^i)f(p^{a-i}) $$ all but the first two summands are zero due to $\mu(p^i) = 0$ for $i\ge 2$. Hence $$ 0 = \sum_{i=0}^a \mu(p^i)f(p^i)f(p^{a-i}) = \mu(1)f(1)f(p^a) + \mu(p)f(p)f(p^{a-1}) $$
2. Induction on $a$.