How to prove that $C_1e^x$ is the unique solution to $f'=f$? Is it possible to prove that $C_1e^x$ is the unique solution to $f'(x)=f(x)$?
I have tried to suppose there exists $g'=g$ and $g(x)\neq C_1e^x$. But I cannot find any contradiction by myself.
Any solutions or hints will be helpful. Thanks.
 A: You can't have that $C_2$ there. But you know that $e^x$ is one solution. If $f$ is another, then $$\frac{\rm d}{{\rm d}x}\frac{f(x)}{e^x} = \frac{f'(x)e^x - f(x)e^x}{e^{2x}} = \frac{f(x)-f'(x)}{e^x} = 0,$$hence $f(x)/e^x = c$, and so $f(x) = ce^x$, for some $c \in \Bbb R$.
A: Suppose $f$ is a solution. Then let $\phi(x) = e^{-x} f(x)$ and note that
$\phi'(x) = 0$ and hence $\phi$ is a constant. Hence
$f(x) = \phi(0) e^x$. 
Hence the solution is unique (modulo the initial value).
A: If $f(x)>0$, $f'(x)>0$ and $f$ is growing, hence remains positive. Conversely $f(x)<0$ remains negative. $f$ cannot vanish unless it is identically zero ($f(x)=0$).
Then
$$\frac{f'(x)}{f(x)}=\left(\log|f(x)|\right)'=1$$ and $$|f(x)|=e^{x+c}=Ce^x.$$

Alternatively, let $f,g$ be two solutions. Then
$$(\log f-\log g)'=\frac{f'}f-\frac{g'}g=0$$ so that $$\frac fg=C.$$
A: The following proof is equivalent to the proofs of IvoTerek and copper.hat, but formulated a bit different.
Start with the differential equation $f' = f$. Rewrite it as $f' - f = 0$ and multiply with $e^{-x}$ (the integrating factor):
$$e^{-x} f' - e^{-x} f = 0$$
The left hand side can now be written as the derivative of a product:
$$e^{-x} f' - e^{-x} f = \left( e^{-x} f \right)'$$
Thus we have the differential equation $\left( e^{-x} f \right)' = 0$. Taking the antiderivative gives $e^{-x} f = C$ for some constant $C$. This gives
$$f(x) = C e^x$$
