Establish the inequality $e^x \geq 1-\ln (1-x)$ for $x \leq 0$. I'm very confused as to where to start on this one.  I assume I would start with some smaller inequalities then work towards this one with algebraic manipulation but I can't figure out where to start.
Define 
$$
f(x)=e^x-1+\ln(1-x)
$$
where we hope to show $f(x)\geq 0$ for $x\leq 0$. Note that at $x=0$ we have 
$$
f(0)=1-1=0
$$
And we have that 
$$
f'(x)=e^x-\frac{1}{1-x}
$$
where we note that 
$$
f'(x)=e^x-\frac{1}{1-x}\leq 0\iff e^x\leq \frac{1}{1-x}\stackrel{x\leq 0}{\iff} e^{-x}\geq1-x
$$
a well known equality derived from the mean value theorem or visual obvious-ness. Thus your function is decreasing to $0$ on the left of the $y$ axis.
PLEASE NOTE THERE IS AN ISSUE WITH THIS ANSWER AT THE SERIES EXPANSION STAGE. I CAN'T DELETE IT AS IT WAS ACCEPTED. - Karn

Hint: 

Let $f(x)=e^x$ and $g(x)=1-\ln(1-x)$.

We know that $f(0)=g(0)=1$.

We need $f(x)\ge g(x)$ so if $f(x)-g(x)$ is positive and decreasing then our proof is done. You may want to consider the Taylor expansion to help with the proof.

Following this is the full solution but I suggest you do not look at it until you have attempted it yourself.


   We have $f'(x)=e^x$ and $g'(x)=\frac{1}{1-x}$.\begin{align}f'(x)-g'(x)&amp;=e^x-\frac{1}{1-x}\\&amp;=\sum_{r=0}^\infty{\frac{x^r}{r!}}-\sum_{r=0}^\infty{x^r}\\&amp;&lt;0\end{align}