# Prove that $\frac{1}{1-x}\geq e^x$ for all $x<0$

I need to prove that $\frac{1}{1-x}\geq e^x$ for all $x<1$. I've allready proved that $1+x\leq e^x$ for all $x\in\mathbb{R}$. I've had some other ideas but I'm not allowed to use anything to different from basic analysis of functions and series, i.e. no differentials or integrals.

Let $y=-x$
$$\frac{1}{1-x}\geq e^x \iff \frac{1}{1+y}\geq e^{-y}\iff e^y \geq 1+y$$
You have $1+x\leq e^x$ so $\dfrac{1}{1+x}\geq e^{-x}$, let $y=-x<0$ then $$e^y\leq\dfrac{1}{1-y}$$