prove or disprove $$\sum_{n=k}^{\infty}\binom{n}{k}\left(\dfrac{-z}{1-z}\right)^n= (1-z)(-z)^k$$
my try: since $$\binom{k}{k}\left(\dfrac{-z}{1-z}\right)^k+\binom{k+1}{k}\left(\dfrac{-z}{1-z}\right)^{k+1}+\binom{k+2}{k}\left(\dfrac{-z}{1-z}\right)^{k+2}+\cdots=\left(\dfrac{-z}{1-z}\right)^{k}\left[\binom{k}{k}+\binom{k+1}{k}\dfrac{-z}{1-z}+\binom{k+2}{k}\left(\dfrac{-z}{1-z}\right)^2+\cdots\right] =\left(\dfrac{-z}{1-z}\right)^k\left[\binom{k}{0}\left(\dfrac{-z}{1-z}\right)^0+\binom{k+1}{1}\dfrac{-z}{1-z}+\binom{k+2}{2}\left(\dfrac{-z}{1-z}\right)^2+\cdots\right] $$