An entire function $f$ satisfying $|f'(z)|\leq |f(z)|$ takes the form $f(z)=\beta e^{\alpha z}$ where, $|\alpha|\leq 1$. Let $f:\Bbb{C}\rightarrow\Bbb{C}$ be an entire function such that $|f'(z)|\leq|f(z)|$ for all $z\in\Bbb{C}$. Show that, there exists $\alpha,\beta\in\Bbb{C}$ with $|\alpha|\leq 1$ such that  $f(z)=\beta e^{\alpha z}$ for all $ z\in\Bbb{C}$.
I am kind of hoping that, we have to take $g(z)=\log(f(z))$ in $\mathbb{C}\setminus\{0\}$. Then the given condition gives $|g'(z)|\leq1$. Then using Liouville's theorem, $g'(z)$ is constant function, equals to \alpha and so on. But there is a flaw in this approach, that is Liouville's theorem can not be used, as $g(z)$ is not defined on whole $\Bbb{C}$.
Please help me with this. Thank you.

We may assume $f$ is not identically $0$ (since $0$ does satisfy the condition, with $\beta=0$).
Suppose $f$ has a zero at $p$.  If this has order $1$, $f'$ has no zero there, and $|f'(p)|>|f(p)|$.  If it has order $m > 1$, $f'$ has order $m-1$, and $\lim_{z \to p} |f'(z)|/|f(z)| = \infty$.
Conclude $f$ has no zeros, and use Liouville on $f'/f$.
