Use induction to show that:
(1+x) E (n, i=0) (-x)^i = 1 + x^(n+1)
The E (n, i=0) bit is the summation notation with the n on top and i=0 on the bottom, you know?
I assume you meant you want to show $\displaystyle \displaystyle (x+1)\sum_{i=0}^n (-x)^i=(-1)^nx^{n+1}+1 $
I leave the base case to you.
Suppose your assertion is true for $\displaystyle n $ i.e. $\displaystyle \displaystyle (x+1)\sum_{i=0}^n (-x)^i=(-1)^nx^{n+1}+1 $
Now $\displaystyle \displaystyle (x+1)\sum_{i=0}^{n+1} (-x)^i=(x+1)(-x)^{n+1}+(x+1)\sum_{i=0}^{n+1} (-x)^i=(x+1)(-x)^{n+1}+(-1)^nx^{n+1}+1 $
Can you finish up from here?