Entire function non identically zero implies that limit sequence of zeros diverges

Let $$f: \mathbb{C} \rightarrow \mathbb{C}$$ an entire function and let $$(a_n)_{n\geq 1} \subset \mathbb{C}^{*}$$ the sequence of the zeros of $$f$$. Suppose that $$z=0$$ is a zero of $$f$$ of order $$m\geq 1$$ and that $$f$$ has infinitely many zeros. Show that:

\begin{align} f \not\equiv 0 \quad \Longrightarrow \quad \lim_{n\to\infty} = |a_n| = \infty \end{align}

I tried to prove it by contradiction using the identity principle for holomorphic functions, without success.

Any suggestions? Thanks in advance!

(It’s a step in the proof of Weierstrass factorization theorem)

Otherwise, $$(a_n)_{n\in\mathbb N}$$ has a bounded subsequence. So, by the Bolzano-Weierstrass, it has a convergent subsequence. But, since $$f$$ is not the null function, this is impossible, by the identity theorem.
Assume that $$\lim_{n\to\infty} |a_n| = \infty$$ does not hold. Then there is a $$R> 0$$ and a subsequence $$(a_{n_k})$$ such that $$|a_{n_k}| \le R$$ for all $$k$$. The closed disk $$D = \{ z : |z| \le R \}$$ is compact, therefore $$(a_{n_k})$$ has an accumulation point. So $$f$$ and the zero function have the same values on a set with an accumulation point in $$\Bbb C$$, and the identity theorem says that $$f \equiv 0$$, contrary to the assumption.