If $f'(x)=1+f(x)$ and $f(0)=0$ prove that $f(x) = e^x -1$

As already told in the previous answer, this can be solved by just dividing $\ f(x)+1 $ and then integrating. One more method comes to mind: $$\ f'(x)=f(x)+1$$ $$\ f'(x)-f(x)=1$$
$$\ {e}^{-x}f'(x)-{e}^{-x}f(x)={e}^{-x}$$ $$\ \frac{d}{dx} {e}^{-x}f(x)={e}^{-x}$$ $$\ {e}^{-x}f(x) = -{e}^{-x}+C$$ $$\ f(x)=C{e}^{x}-1 $$ Find $\ C=1$ by using $\ f(0)=0$. Hope it helps!

EDIT: This is a general trick(manipulation) used that whenever there is a $\ f'(x)+Kf(x)$, multiply and divide by $\ {e}^{Kx} $ so that the expression reduces to $$\ {e}^{-Kx}\frac{d}{dx} {e}^{Kx}f(x)$$

So you have $${dy\over dx} =1+y$$ Rewrite it like this $${dy\over 1+y} = dx$$ and now integrate both sides.