Let $f$ be a multiplicative function satisfying $\lim_{p^m\to \infty} f(p^m)=0$. Show that $\lim_{n\to\infty} f(n)=0$.
By unique factorization, we can write $n=\prod_{i=1}^k p_i^{\alpha_i}$, where $p_1,\dots,p_k$ are distinct primes. I get stuck here. If $n$ is sufficiently large, can we say that there exits an index $i$ such that $p_i^{\alpha_i}$ is sufficiently large?
since $|f(p^m)| \to 0$ when $p^m \to \infty$, there exists $M$ such that $p^m > M \implies |f(p^m)| < 1$.
this proves that for every $n$ : $$|f(n)| < \prod_{p^m < M} |f(p^m)| = C$$
now, use the lemma that for every $K$, there exists $N(K)$ such that $$n > N(K) \qquad \implies \qquad \exists p^m > K, \qquad p^m | n$$
finally, for any $\epsilon > 0$, find $K(\epsilon)$ such that $p^m > K(\epsilon) \implies |f(p^m)| < \epsilon$,
and for any $n > N(K(\epsilon))$, let $p^m$ be the greatest prime power dividing $n$,
by the lemma we know that $p^m > K(\epsilon)$, we also have $n = p^m \frac{n}{p^m}$ such that $gcd(p^m,\frac{n}{p^m}) = 1 $ hence : $$ f(n) = \underbrace{f(p^m)}_{< \ \epsilon} \underbrace{f(\frac{n}{p^m})}_{< \ C} \qquad \implies \qquad |f(n)| < C \epsilon$$
i.e. $$\lim_{n \to \infty} f(n) = 0$$