If $f$ is entire and $|f|\geq 1$, then show $f$ is constant.
If $f$ is holomorphic at $z_0$ and $f(z_0)$ is nonzero, then $\frac{1}{f}$ is holomorphic at $z_0$, and from here you can conclude.
Let $g(z) = \frac{1}{f(z)}$. Then, since $|f(z)| \geq 1$ for all $z$ and is entire, therefore so is $g$ and moreover, we have that $|g(z)| \leq 1$ for all $z$.
Hence by Liouville's theorem, $g$ is bounded, but then so is $f$.