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.

You can work directly with properties of the binomial coefficient. For $k\ge 0$ let $$f_k(x)=\sum_{n\ge 0}\binom{n}kx^n\;.$$ Then
$$\begin{align*}
f_k(x)&=\sum_{n\ge 0}\binom{n}{k}x^n\\
&=\sum_{n\ge 0}\left[\binom{n-1}{k-1}+\binom{n-1}{k}\right]x^n\\
&=\sum_{n\ge 0}\left[\binom{n}{k-1}+\binom{n}k\right]x^{n+1}\\
&=x\sum_{n\ge 0}\binom{n}{k-1}x^n+x\sum_{n\ge 0}\binom{n}kx^n\\
&=xf_{k-1}(x)+xf_k(x)\;,
\end{align*}$$
so $(1-x)f_k(x)=xf_{k-1}(x)$, and $$f_k(x)=\frac{x}{1-x}f_{k-1}(x)\;.\tag{1}$$
Since $$f_0(x)=\sum_{n\ge 0}\binom{n}0x^n=\sum_{n\ge 0}x^n=\frac1{1-x}\;,$$ an easy induction yields the desired result.
