Trouble with form of remainder of $\frac{1}{1+x}$ While asking a question here I got the following equality
$\displaystyle\frac{1}{1+t}=\sum\limits_{k=0}^{n}(-1)^k t^{k}+\frac{(-1)^{n+1}t^{n+1}}{1+t}$
I'm trying to prove this with Taylor's theorem, I got that $f^{(n)}(x)=\displaystyle\frac{(-1)^nn!}{(1+x)^{n+1}}$ so the first part of the summation is done but I can't seem to reach the remainder according to Rudin the form of the remainder is  $\displaystyle\frac{(-1)^n!}{(1+c)^{n+1}}t^{n+1}$ . I know the equality can be proved by expanding the sum on the right but I want to get the result with the theorem. 

Taylor will not help you here.
Hint: $\sum_{k=0}^n r^k=\frac{1-r^{n+1}}{1-r}$ for all positive integers $n$, when $r\neq 1$.
Take $r=$...
