Problem: Suppose that $f$ is an entire function and that there is a bounded sequence of distinct real numbers $\{ a_n \}_{n\in \mathbb{N}}$ such that $f(a_n)\in \mathbb{R}$ for all $n\in \mathbb{N}$. Show that $f(x)\in \mathbb{R}$ for all $x\in \mathbb{R}$.
My attempt: Since $\{ a_n \}$ is a bounded sequence of real numbers, there exists a convergent subsequence $\{ a_{n_k} \} \rightarrow a \in \mathbb{R}$. By $f$ entire, we can consider its Taylor expansion at $a$: $$f(z) = \sum\limits_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (z-a)^n$$ If we can show that $f^{(n)}(a) \in \mathbb{R}$ for all $n\in \mathbb{N}$ then we are done since all the terms are real so plugging in a real number $x$ into $f(z)$ will output a real value.
Since $f$ is entire, it is a continuous function so $f(a) = f(\lim\limits_{k \rightarrow \infty} a_{n_k}) = \lim\limits_{k \rightarrow \infty} f(a_{n_k}) \in \mathbb{R}$ since $\mathbb{R}$ is a closed subset of $\mathbb{C}$ and $f(a_{n_k}) \in \mathbb{R}$ by assumption. This implies that $$f'(a) = \lim\limits_{z\rightarrow a} \frac{f(z) - f(a)}{z-a} = \lim\limits_{k\rightarrow \infty} \frac{f(a_{n_k}) - f(a)}{a_{n_k}-a} \in \mathbb{R}$$ since $f(a_{n_k}), f(a), a_{n_k}, a \in \mathbb{R}$.
Question: How can I show that $f^{(n)}(a) \in \mathbb{R}$ for $n \geq 2$ from here? I have not used the "distinct real numbers" assumption yet.