So I have $f(x)=e^x-1-x$, MVT gives me $f'(c)={f(x)-f(0)\over{x-0}}={e^x-x-1\over x}$. So $f(x)=xf'(c)$. I got stuck at how exactly I can explain that $f(x)\ge 0\, \forall\, x\ge0$
I know that $f'(c)$ has the same sign as $c$ which has the same sign as $x$, thats why $f'(c)\ge0$ when $x\ge0$, which makes $f(x)\ge0$, but how do I show it nicely?