Proving $f(x)>0$ on $\mathbb{R}^+$ given that $f(0)=0$ and $f'(x)>f(x)$ Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be differentiable, with the properties
(1)  $\ f(0)=0$, and
(2) $\ f'(x)>f(x)$ for all $x\in\mathbb{R}.$
Prove that $f(x)>0$ for all $x\in\mathbb{R}^+.$
It's clear that one can get that $f$ is positive on some interval, by utilizing the $\epsilon$-$\delta$ definition of the derivative (using $\epsilon=f'(0)/2>f(0)/2=0$) for all positive $x$ in some $\delta$-neighborhood centered at $0$. How would one get this over all positive reals? I've tried to proceed by contradiction, but that hasn't yielded much for me thus far, using things like the mean value or creating auxiliary functions. Maybe someone with a bit more wit could direct me down the right path?

Set $g(x) = e^{-x}f(x)$. One then has $g(0) = 0$ and $g'(x) = -e^{-x}f(x) + e^{-x}f'(x) = e^{-x}(f'(x)-f(x)) > 0$ for all $x$. 
Thus, $g(x) > g(0)$ for all $x \in \mathbb{R}^+$, or $f(x) > 0$ for all $x\in \mathbb{R}^+$.
