Why does $\sum_n\binom{n}{k}x^n=\frac{x^k}{(1-x)^{k+1}}$? I don't understand the identity $\sum_n\binom{n}{k}x^n=\frac{x^k}{(1-x)^{k+1}}$, where $k$ is fixed.
I first approached it by considering the identity
$$
\sum_{n,k\geq 0} \binom{n}{k} x^n y^k = \sum_{n=0}^\infty x^n \sum_{k=0}^n \binom{n}{k} y^k = \sum_{n=0}^\infty x^n (1+y)^n  = \frac{1}{1-x(1+y)}.
$$
So setting $y=1$, shows $\sum_{n,k\geq 0}\binom{n}{k}x^n=\frac{1}{1-2x}$. What happens if I fix some $k$ and let the sum range over just $n$? Thank you.

Start from the Taylor series of $\frac1{1-x}$, differentiate it $k$ times, and then multiply everything by $\frac{x^k}{k!}$.
