Prove $f'(z)=f(z)$ implies $f(z)=ce^z$ for some $c \in \mathbb{C}$ Suppose that: $$f'(z)=f(z) \text{ for all }z\in\mathbb C.$$ In other words, the complex function $f$ is equal to its own derivative.  Prove that there is a constant $c\in\mathbb C$ such that $f(z)=c e^z$; that is $f$ is the exponential function (up to a multiplicative constant).

So far, I've tried substituting $f'(z)=f(z)$ into the limit definition of $f'(z)$, to no avail. I'm trying to think of what other simple expressions I have relating $f$ and $f'$, but am not having much success.
 A: Let $g(z) = f(z) e^{-z}$. It's the product of two differentiable functions, hence it is differentiable. By the product rule, $g'(z) = f'(z) e^{-z} + f(z) (-e^{-z}) = 0$. Hence $g' = 0$, so $g$ is a constant, say $g(z) = c$. Therefore $f(z) = c e^{z}$.
Note that a priori, you do not know that the complex logarithm of $f$ is well defined. And in fact it's possible that $f = 0$, so in that case it's really not defined at all...
A: Re-write as
\begin{equation}
\frac{df}{dz} = f
\end{equation}
Thus
\begin{equation}
\frac{df}{f} = dz
\end{equation}
from which (and wlog)
\begin{equation}
\int \frac{df}{f}=\int dz
\end{equation}
Hence,
\begin{equation}
\ln f=z+c
\end{equation}
Hence 
\begin{eqnarray}
f(z)&=&e^{z+c} \\
    &=& Ce^{z}
\end{eqnarray}
Where $C=e^{c}$
-----------EDIT-----------------------------------------------------------------
Actually, due to comments below and having another look at the question there is nothing regarding $f$ and how well $f$ is defined. Consequently, one could consider
\begin{equation}
\frac{d}{dz}\left(e^{-z}f(z)\right) = -e^{-z}f(z)+f'(z)e^{-z}
\end{equation}
Clearly substituting our condition $f'(z)=f(z)$ we arive at
\begin{equation}
f(z)(e^{-z}-e^{-z}) = 0
\end{equation}
Consequently
\begin{equation}
e^{-z}f(z)=C
\end{equation}
where $C$ is a constant.
Thus
\begin{equation}
f(z)=Ce^{z}
\end{equation}
A: Hint: note that
$$
\frac{d}{dz} \ln[f(z)] = \frac{f'(z)}{f(z)} = 1 
$$
A: At first you should show that the function is holomorphic (by using Cauchy-Riemann differential equations).
Then, from a theorem in complex analysis, above differential equation can be solved by the separation of variables.
To show the other direction, simply substitute the function $f(z) = ce^z$ into the equation $f'(z) = f(z)$.
