Prove that f'=f iff f is an exponential funtion Written more formally, prove that $f' = f \iff \exists c \in \mathbb{R} : f = c * \exp$
In other words, I guess, it's enough to prove that $\exp$ and $f(x) = 0$ are the only functions that are equal to its derivatives. How can I do that? I'll be grateful for a hint instead of a full proof.
Thanks!
 A: For a hint, consider the function $h(x)=f(x)e^{-x}$ where $f(x)$ is a solution of the differential equation. 
However, for the sake of completion, I shall write out the full proof. Do not read it if you do not wish, unless you get stuck.
$f(x)=ce^x$ then $f'(x)=ce^x$ since the derivative of $e^x$ is $e^x$ and $c$ is a constant. Thus $f'(x)=f(x)$
Thus, we can easily see that $f(x)=ce^x$ satisfies the differential equation, and thus we can safely say that a solution exists.
$h'(x)=f'(x)e^{-x}-f(x)e^{-x}=(f'(x)-f(x))e^{-x}=0$ since $f(x)=f'(x)$ is the same as $f'(x)-f(x)=0$
Thus, $h(x)=c$ where $c$  is an arbitrary constant.
Therefore, $f(x)=ce^x$.
QED.
A: Using a simple differential equation and taking a derivative:
Proof:
First, suppose $f(x) = y$ and $\frac{dy}{dx} = y$. Then $dy = y \hspace{1mm} dx \Rightarrow \frac{1}{y} dy = dx \Rightarrow ln(y) = x + C_{1} \Rightarrow y = Ce^{x} $
Thus, we see that if $f'(x) = f(x)$, then $f = Ce^{x}$.
Conversely, if $f(x) = y = Ce^{x}$, then we have that $\frac{dy}{dx} = \frac{d}{dx}Ce^{x} = C \frac{d}{dx} e^{x} = Ce^{x}$
Thus, we see that if $f(x) = Ce^{x}$, then $f' = f$.
This completes the proof.
