Complex inequalities and constants

Question

Let $f(z)$ be an analytic function. Show that if $|f(z)| > 1 + |e^z|$ then $f(z)$ is constant.

I have no idea what to do, I've subbed $z= x+iy$ and got $|f(z)|>1+e^x$ but lost here.

Answer 1

Hint: Clearly $f$ has no zeros. Then

$$\left|\frac{1 + e^z}{f(z)}\right| < 1$$ for all $z \in \mathbb{C}$. What can you conclude now?

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Alternatively, if you prefer even stronger theorems, note that $f(\mathbb{C})$ is missing two points.