Consider the polynomial $f_{k}\left ( x \right )=\sum_{i=0}^{k}x ^{i}$. Find ${f_{k}}'\left ( 1 \right )$ in terms of $k$.
If I think of $f_{k}\left ( x \right )=\sum_{i=0}^{k}x ^{i}$ as $x^{0}+x^{1}+x^{3}+... +x^{k}$, it's derivative would be $1+3x^2+4x^3+ ... +kx^{k-1}$ and ${f_{k}}'\left ( 1 \right )$ would equal $\frac{k^{2}+k}{2}$. However, if I think of $f_{k}\left ( x \right )=\sum_{i=0}^{k}x ^{i}$ as $\frac{1-x^{k+1}}{1-x}$, it's derivative would be $\frac{(x-1)(k+1)x^{k}+(1-x^{k+1})}{(1-x)^{2}}$ (quotient rule) and ${f_{k}}'\left ( 1 \right )$ would be ... $\frac{0}{0}$?
$$f_{k}\left ( x \right )=\sum_{i=0}^{k}x ^{i}=1+x+x^2+x^3+...+x^k$$
$$ f'_{k}\left ( x \right )=1+2x+3x^2+4x^3+...+kx^{k-1}=\sum_{i=1}^{k} ix ^{i-1} $$
$$f'_{k}\left ( 1 \right )=1+2+3+4+...+k= \frac {k(k+1)}{2}$$
You can not write your function as $$\frac{1-x^{k+1}}{1-x}$$ around $x=1$ because it is not defined at $x=1$
Same with your derivative,which is not defined as $x=1.$
Have you tried to simplify your derivative and see what is the result after removing the $(1-x)^2 $ from the top and bottom?