Prove that if $f=f'$ then $f$ is monotone. Suppose I do not know that $e^x$ solves the equation $$f'(x)=f(x),\;\;\;x\in\mathbb{R}.$$ I am just given this equation and want to see if $f$ is increasing. Is there a way to prove that $f$ is increasing without going through power series (with all theorems it requires) and redefinig the exponential function?
EDIT 1
I meant monotone not increasing.
 A: Answer for the earlier version of the question: $f(x)=-e^{x}$ also satisfies this property so $f$ need not be increasing. 
PS $f^{2}$ has derivative $2ff'=2f^{2} \geq 0$ so $f^{2}$ is increasing. It follows that $f$ is monotonic (see for example If $f^2$ monotonically increasing in $R$ then f monotonic).
A: Assume that $f'$ has not the same sign everywhere, i.e. $f'(x_1) > 0$ and $f'(x_2) < 0$. 
If $x_1 < x_2$ then $f$ attains its maximum on $[x_1, x_2]$ at a point $x_3$ in the open interval $(x_1, x_2)$.  Then $f(x_3) \ge f(x_1) = f'(x_1) > 0$, but $f'(x_3) = 0$.
If $x_2 < x_1$ then the same argument is used with the minimum on $[x_2, x_1]$, giving also a contradiction.
A: It's easy to see $f_0(x) = 0$ is a fixed point of the derivative, so if $f(x)$ satisfies $f'(x) = f(x)$ and $f(0) > 0,$ then $f'(x) = f(x) > 0$ for all $x$.  Similarly, if $f(0) < 0$, then $f'(x) = f(x) < 0$ for all $x$.
A: One can prove that if such a function vanishes once it vanishes everywhere.
If $f(a) =0$ then we prove that $f(x) =0$ for all $x$. Suppose there is a $b\neq a$ with $f(b) \neq 0$ and then let $F(x) =f(b-a+x) f(a+b-x) $. Clearly $$F'(x) = f'(b-a+x) f(a+b-x) - f(b-a+x) f'(a+b-x) =0$$ (via $f'=f$) and hence $F$ is constant. Then $F(x) =F(a)=f(b)f(b)>0$ and $F(b) =f(2b-a)f(a)=0$ and we have a contradiction.
Thus one option is that $f(x) =0$ for all $x$ and another option is that $f$ does not vanish anywhere. By continuity $f$ maintains a constant sign and $f'=f$ shows that $f$ is strictly monotone. 
