If an abelian group contains at most $n$ elements of order divisible by $n$, is it cyclic?

Question

I usually see this question as "if $G$ has at most $n$ solutions to $x^n=1$, then $G$ is cyclic". But here, we have that if $G$ has at most $n$ elements of orders divisible by $n$, then $G$ is cyclic. I am wondering if the following proof works. 3 Let $G$ be such a group with order $|G| = p_1 ^{\alpha_1} \cdots p_k ^{\alpha_k}$ where each $p_i$ is a distinct prime. Every non-identity element of any $P_i \in \text{Syl}_{p_i}(G)$ has order divisible by $p_i$ by Lagrange's theorem, but by hypothesis, there are at most $p_i$ such elements, so $|P_i| \le p_i$. Since $P_i$ is not trivial, this means $P_i \cong C_{p_i}$. Furthermore, $P_i$ is normal because $G$ is abelian. Because this holds for all $i$, we see that $\alpha_i = 1$ and $$ G \cong C_{p_1} \times \cdots \times C_{p_k} \cong C_{p_1 p_2 \cdots p_k} \cong C_{|G|}. $$