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.

The easiest way to see this is to use the common generalization of binomial coefficients with negative upper index, which satisfy
$$
  \binom{-n}k=(-1)^k\binom{n+k-1}k,
$$
and for which the binomial theorem in the form
$$
  (1+X)^{-n}=\sum_{k\geq0}\binom{-n}kX^k
$$
continues to hold, as an identity of formal power series.
Now in order to get your exponents to match the lower index, write
$$\begin{align}
  \sum_{n\geq k}\binom nkX^n
&=\sum_{n\geq k}\binom n{n-k}X^n
=\sum_{m\geq0}\binom{k+m}mX^{k+m}
\\&
=X^k\sum_{m\geq0}(-1)^m\binom{-k-1}mX^m
\quad\text{by above identity}
\\&
=X^k(1-X)^{-k-1}
\quad\text{by the binomial theorem}
\\&
=\frac{X^k}{(1-X)^{k+1}},
\end{align}
$$
an identity of formal power series (which continues to hold when substituting a value for $X$ small enough to make the series on the left converge).
