How to show that $f(p^{k}) = f(p) \cdot f(p^{k-1}) \Longrightarrow f(p^{k}) = [f(p)]^{k}$

Question

If f is an arithmetic function such that $f (1) = 1$ and $p$ is a prime number. Prove that:

$\forall k \in \mathbb{N}$

$f(p^{k}) = f(p) \cdot f(p^{k-1}) \Longrightarrow f(p^{k}) = [f(p)]^{k}$

Answer 1

As suggested in comments, this can be easily shown using induction.

$1^\circ$ For $n=1$ the claim is true, since $$f(p^1)=f(p).$$

$2^\circ$ (inductive step)

Assume that the claim is true for $n-1$, we will prove that it is true for $n$.

$$f(p^n) \overset{(1)}= f(p)f(p^{n-1}) \overset{(2)}= f(p) [f(p)]^{n-1} = [f(p)]^{n}$$ In the equality (1) we have used the given condition. In the equality (2) we have used inductive hypothesis.