# How to prove the closed form for the generating function $g_r(x) = \sum_{n=0}^\infty {n \choose r} x^n$ possibly using Newton's Binomial Theorem

define the generating function $$g_r(x) = \sum_{n=0}^\infty {n \choose r} x^n$$

how do you find the closed form for this function?

I got $$x^r{(1-x)}^{-(r+1)}$$ but I am not sure how to prove it. I think it relates to Newton's Binomial Theorem and I know we have $$(1-z)^{-n} = \sum_{k=0}^\infty {n+k-1 \choose k} z^k$$ where $$|z| < 1$$, but again, not sure about the right steps to take.

any help is appreciated! thank you

Actually you have found all results needed for the proof. It remains only to order them correctly. In what follows I assume that $$r$$ is a non-negative integer.
We have by the extended binomial theorem: $$(1-x)^{-r}=\sum_{k=0}^\infty\binom{-r}k (-x)^k=\sum_{k=0}^\infty\binom{r+k-1}k x^k.$$
Therefore: $$x^r(1-x)^{-r-1}=\sum_{k=0}^\infty\binom{r+k}k x^{r+k}=\sum_{k=0}^\infty\binom{r+k}r x^{r+k}=\sum_{n=r}^\infty\binom{n}r x^{n},$$ where in the last step we reindexed the summation: $$k\mapsto n-r$$.