Prove that $f\equiv 0$ where $f$ is entire 
Problem:
  Let $f$ be an entire function such that $f(0)=f'(0)=0$ and $|e^{f(z)}| \leq e^{|z|}$ for all $z\in \Bbb{C}$. Then $f\equiv 0$

My attempt: I genuinely tried applying all tools I'm familiar with, but I haven't produced anything that seems particularly fruitful. Here are some thoughts I've had:
Since $f$ is entire, $f$ can be written as a power series $f(z)=\sum_{k=0}^{\infty} a_k z^k$, $z\in \Bbb{C}$. The conditions $f(0)=f'(0)=0$ imply that $a_0=a_1=0$. Then the function $g(z)=\frac{f(z)}{z^2}$ $(z\neq 0)$ has a removable singularity in the origin, so without loss of generality, $g$ is entire. Somehow we have to use the estimate $|e^{f(z)}| \leq e^{|z|}$ but I can't figure it out.
I'm thankful for any help.
 A: By the Borel–Carathéodory theorem (https://en.wikipedia.org/wiki/Borel%E2%80%93Carath%C3%A9odory_theorem) and because $\exp\left(\Re\left(f\left(z\right)\right)\right) = |\exp\left(f\left(z\right)\right)| \leq \exp\left(\left|z\right|\right)$ implies $\Re\left(f\left(z\right)\right) \leq \left|z\right|$, we have for every $0 < r \leq R$ the estimate
\begin{equation}
\left|f\left(z\right)\right| \leq \frac{2r}{R-r} R + \frac{R+r}{R-r}\left|f\left(0\right)\right| = \frac{2r}{R-r} R.
\end{equation}
Since the coefficients $a_n$ are give by
\begin{equation}
a_n = \frac{1}{2\pi i}\int_{\left|z\right| = r} \frac{f\left(z\right)}{z^{n+1}} \, dz 
\end{equation}
we can estimate that
\begin{equation}
\left|a_n\right| \leq \frac{1}{r^n}\frac{2r}{R-r}R.
\end{equation}
Now choosing R = 2r, we arrive at 
\begin{equation}
\left|a_n\right| \leq \frac{1}{r^n}\frac{2r}{r}2r = \frac{4r}{r^n} \rightarrow 0
\end{equation}
as $r \to \infty$ for every $n \geq 2$. Therefore $a_n = 0$ for all $n \geq 2$. With $f\left(0\right) = f'\left(0\right)=0$ the result follows. 
A: A minor fix: $f(0)=f'(0)=0$ implies $a_0=a_1=0$. 
Let $h(z)=\sum_{n\geq 3}a_n z^n$. Since $\left|e^{f(z)}\right|\leq e^{|z|}$, we have:
$$\text{Re}(h(z)) \leq |z|-\text{Re}(a_2 z^2) \tag{1}$$
hence  if $a_2\neq 0$, $\text{Re}(h(z))< C$ over an unbounded, simply-connected open subset of $\mathbb{C}$. Liouville's theorem applied to $e^{h(z)}$ then gives that $h(z)$ is constant, hence zero, since $h(z)$ is $z^3$ times an entire function. If $a_2=0$, let:
$$ N = \min\{n\geq 2:a_n\neq 0\} $$
and replace line $(1)$ with:
$$\text{Re}(h(z))\leq |z|-\text{Re}(a_N z^N). \tag{2}$$
In the same way we get $h(z)\equiv 0$, so we just have to deal with the case:
$$ f(z) = a_2 z^2 \tag{3}$$
that is straightforward. In order that $\left|e^{f(z)}\right|\leq e^{|z|}$ holds for any big real number $z$, $\text{Re}(a_2)$ must be zero, but in such a case the inequality cannot hold for some $z=(1\pm i)r$, unless $a_2=0$.
