How prove this sum $\sum\limits_{n=k}^{\infty}\dbinom{n}{k}\left(\dfrac{-z}{1-z}\right)^n$

Question

prove or disprove $$\sum_{n=k}^{\infty}\binom{n}{k}\left(\dfrac{-z}{1-z}\right)^n= (1-z)(-z)^k$$

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

Answer 1

$$\sum_{n=k}^{\infty}\binom{n}{k}\left(\dfrac{-z}{1-z}\right)^n= (1-z)(-z)^k$$ \begin{align*} &\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] \end{align*} since $$\dfrac{1}{(1-x)^{n+1}}=\binom{n}{n}+\binom{n+1}{n}x+\binom{n+2}{n}x^2+\binom{n+k}{n}x^k+\cdots$$ let$x=\dfrac{-z}{1-z}$