$g'-g> 0$ with boundary condition $g(0)=0$ 
Exercise:
Let $g$ be a differentiable function $[0,1]\rightarrow\mathbb{R}$ such that $g'-g>0$ on $(0,1)$ and $g(0)=0$. Prove that $g>0$ on $(0,1]$.

There exists a solution using the function $\exp$:

Solution:
Let $h=\exp(-x)g$. Then $h'$ is strictly positive on $(0,1)$ and $h(0)=0$, whence $h>0$ on $(0,1]$ and therefore $g>0$ on $(0,1]$.


Intuitively, the function $g$ starts at $0$ with the value $0$. From there on, its increasing rate $g'$ stays greater than $g$, hence strictly positive. Thus the function $g$ can never decrease.

My question is, does there exist an argument which clarifies the above intuition? (an alternative to the function $\exp$).
 A: A priori, $g'$ and $g$ can be both negative, so solely by intuition may not be enough. However one can get a proof without using the exponential function as follows.
Proving by contradiction, assume that there exists $a_0\in (0,1]$ such that $g(a_0)\leq  0.$ Then by mean value theorem, $$\exists a_1\in (0,a_0)~{\rm such~that~}g'(a_1)=\frac{g(a_0)-g(0)}{a_0-0}=\frac{g(a_0)}{a_0}\leq 0.$$ But $$g'(a_1)>g(a_1)$$$$\Rightarrow g(a_1)<g'(a_1)=g(a_0)/a_0\leq 0$$ $$\Rightarrow g(a_1)<g(a_0)\leq0~(\because 0<a_0\leq 1),$$ which shows that the absolute minimum of $g$ on $[0,a_0]$ occurs at a point in $(0,a_0)$. Let $\alpha$ be such a point, so $g(\alpha)$ is the absolute minimum of $g$ on $[0,a_0]$. By mean value theorem again, there exists $\beta\in (0,\alpha)$ such that $$g'(\beta)=\frac{g(\alpha)-g(0)}{\alpha-0}=\frac {g(\alpha)}{\alpha}$$ and $$g'(\beta)-g(\beta)>0$$ $$\Rightarrow g(\beta)<g'(\beta)=\frac{g(\alpha)}{\alpha}<g(\alpha)~(\because 0<\alpha<1,g(\alpha)<0).$$ But $g(\beta)<g(\alpha)$ gives a contradiction that $g(\alpha)$ is the absolute minimum of $g$ on $[0,a_0]$. One concludes that $g(x)>0,\forall x\in (0,1].$ QED
