Bounded away from the origin implies constant

Suppose that $f$ is entire and there is a positive number $M$ such that $|f(z)|\geq M$ for all $z\in\mathbb C$. Prove that $f$ is constant.

This is problem 13 from chapter 5.

Suppose that f is not constant, thus it exists $z_o$ such that $f'(z_o) \neq 0 $

let R $>$ 0 and consider the circle C[$z_o$,R], since f entire then 1/f is holomorphic on an open set that contains C then we can apply Cauchy theorem on C

$\frac{1}{f'(z_o)}=\frac{1}{2i\pi}\int_{C} \frac{1/f(z)}{(z-z_o)^2}dz$. then we have:

$\left|\frac{1}{f'(z_o)}\right| <= \frac{1}{2\pi}$ $\left|\frac{1/f(z)}{(z-z_o)^2}\right|$.length C[$z_o$,R]
$<= \frac{1}{2\pi}\left|\frac{1/f(z)}{R^2}\right| 2\pi R$
$<=\frac{1}{M R}$

if we make R goes to $\infty$ we get $\left|\frac{1}{f'(z_o)}\right|=0$ contradiction!

Thus f is constant.