A multiplicative function in Number theory that is $0$ for all sufficiently large enough $n\in\mathbb{N}$ Dear MSE Mathematicians , I have a query regarding proof verification of the following Theorem
$\textbf{Theorem}$ Suppose that $f$ is a multiplicative function such that $$\lim_{p^{k}\rightarrow \infty} f(p^k)=0$$ Where $p^k$ denote the prime powers.
Then $$\lim_{n\rightarrow \infty} f(n)=0$$
$\textbf{Proof}$
Since $$\lim_{p^{k}\rightarrow \infty} f(p^k)=0$$ there are only finitely many prime powers such that $|f(p^k)|\geq 1$. Among all such prime powers choose the largest since it's finite there exist one largest say $q^j$ Then , Define $$n=\prod_{ \text{atleast one prime power dividing n}>
 q^j,p|n}p^{v_p(n)}$$.
$\textbf{Claim}:$ There exist a positive integers $N$ such that every integer $>N$ takes the form of $n$.
Proof: Define $$S=\{n|\text{atleast one prime power that divides n is}>q^j\}$$ Look over $S^c$ then clearly every member of $S^c$ is divisible by a a prime power less than $q^j$ so $S^c$ must contains a largest element name that element $N$.
Back to the problem Note that $N+i\in S$ and so every element $a\in A=S\cap \{N+i|i\in\mathbb{N}\}$  has $|f(a)|=0$
implying,
$$\lim_{n\rightarrow \infty}f(n)=0$$
$\blacksquare$
Thanks to Everyone, who provide suggestions and comment for this post.
$\textbf{My Attempt 2}:$
Define $a(p,k)=p^k$ Since we have that $\lim_{p^k\rightarrow \infty} f(p^k)=0$ we have Then there exist $a(N,M)$ such that for all $q>N$ and $j>M$ such that $a(q,j)=0$. Now applying the above Claim in attempt 1. Can we thru?
Is my attempt 2 correct.
Both attempt are wrong.

Note that $\lim f(p^k)=0$ does not imply that only finitely many $|f(p^k)|\ne0$!
To mend the proof, consider $\epsilon>0$.
From the existence of $\lim f(p^k)$, we see that $|f(p^k)|\ge1$ only for finitely many $p^k$. Let $N$ be the product of all $p^k$ with $|f(p^k)|\ge1$ and let $M$ be the product of the corresponding $|f(p^k)|$.
Find $A$ with $|f(p^k)|<\epsilon/M$ for all $p^k>A$. For sufficiently large $n$, at least one prime power dividing $n$ is $>A$ so that $|f(n)|<M\cdot \epsilon/M$.
