If $f$ is entire and monotone on $\mathbb{R}$, then it's a polynomial or $f(1/z)$ has an essential singularity

Question

In the answer to this post, the following fact seems implicitly used:

Suppose $f: \mathbb{C} \to \mathbb{C}$ is entire. Further, suppose $f$ is a bijection $\mathbb{R} \to \mathbb{R}$ and $z$ is real $\iff$ $f(z)$ is real. Then either $f(z)$ is a polynomial, or $f(1/z)$ has an essential singularity at $z=0$.

Is this true? I only learned about essential singularities a few days ago, so I'm lacking intuition.

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My work

Certainly, polynomials may have the stated property; and if $f$ is a polynomial, $f(1/z)$ has a pole at zero.

Now suppose $f$ is not a polynomial. Since $f$ is entire, it equals its Laurent series at $0$, which is not a polynomial. We want to say something about $f(1/z)$.

Entirety of $f$ $\implies$ $f$ can only have non-negative order terms in its Laurent series. As $f$ is not a polynomial, there are infinitely many such terms. Then $f(1/z)$ has infinitely many negative-order terms. This implies $f(1/z)$ has an essential singularity. (This is something I know from lectures.)

Is my proof sound?

Answer 1

If $f$ is not a polynomial function, then its Taylor series centered at $0$ is of the type $\sum_{n=0}^\infty a_nz^n$ with infinitely many $a_n$'s different from $0$. Therefore,$$f\left(\frac1z\right)=a_0+\frac{a_1}z+\frac{a_2}{z^2}+\cdots$$Since $a_n\ne0$ in infinitely many cases, $f\left(\frac1z\right)$ has an essential singularity at $0$.

Note that the fact that $f$ is entire is almost not needed. All that it is needed is that $f$ is a holomorphic function whose domain is a connected open subset of $\Bbb C$ to which $0$ belongs and that $f$ is not a polynomial function.