$f$ is entire, prove that $\{f_n = f(nz) | n \in \mathbb{N}\}$ is normal on the annulus iff $f$ is constant

I am studying for my exam and came across this question:

Suppose $f$ is entire and $r<R$. Prove that the family $\mathcal{F} = \{f_n = f(nz) | n \in \mathbb{N}\}$ for $z \in \mathbb{C}$ is normal on the annulus $r< |z|<R$ iff $f$ is constant.

This is my attempt:

$\Leftarrow$ If $f$ is constant, then $f(z) = z_0, \, \forall z \in \mathbb{C}$. Then since $f_n = f$ for all $n \in \mathbb{N}$. Thus $\mathcal{F} = \{f\}$, and therefore every sequence of $\mathcal{F}$ is the sequence of function $f$, and therefore wil converge uniformly on every compact subset, and thus is a normal family. Or I could also use Montels theorem and see that since $f_n(z) = f(z) = z_0$ for all $n \in \mathbb{N}$ and all $z \in \mathbb{C}$, we know that $\mathcal{F}$ is uniformly bounded by $z_0$ and therefore it is a normal family.

$\Rightarrow$ We know that $f$ is entire, therefore we only need to prove that it is bounded (and then can use Louiville's theorem to conclude that it is constant). Let $D_k(0)$ be a disk, then since it is compact we know that there exists a $B_k$ such that $|f_n(z)| \leq B_k$ for all $z \in \mathbb{C}, n \in \mathbb{N}$. I want to use the following theorem:

Corollary 4.6 Stein and Shakarchi: Suppose that $\Omega$ is a region with compact closure $\overline{\Omega}$. If $f$ is holomorphic on $\Omega$ and continuous on $\overline{\Omega}$ then $$sup_{z\in\Omega}|f(z)| \leq sup_{z \in \delta \Omega} |f(z)|.$$

But I am not sure how to since I don't know if on $D_R(0)$ it holds that $\mathcal{F}$ is uniformly bounded on $D_R$ since I am only given it is normal on $r<|z|<R$.

Can someone tell me if my attempts are correct and how I should go further in my second attempt?

First part is correct. Montel's Theorem is not required.

If $(f_n)%$ is normal then the sequence is uniformly bounded on any compact subset of the annulus. If $r<r_1<R_1<R$ then there exists $M$ such that $|f_n(z)| \leq M$ for $r_1 \leq |z|\leq R_1$. For any $z$ such that there the interval $(\frac {|z|} {R_1}, \frac {|z|} {r_1})$ has length $ >1$ pick an integer $n$ in it. It follows that $|f(z)|=|f(n \frac {z} n) |\leq M$ since $\frac z n \in [r_1,R_1]$. Conclude that $f$ is bounded, hence constant by Liouville's Theorem.