Closed form for $\sum_{n=0}^{\infty} \binom{n+k}{k} x^n$ (Negative Binomial Theorem) I was wondering if there is also a closed expression for the series 
$$\sum_{n=0}^{\infty} \binom{n+k}{k} x^n$$ where $|x|<1.$
A few examples suggest that the answer is $\frac{1}{(1-x)^{k+1}}$ but I don't see how to show this.
 A: You can also proceed by adding a variable. Let:
$$f(x,y) := \sum_{k=0}^{+ \infty} y^k \sum_{n=0}^{+ \infty} \binom{n+k}{k} x^n.$$
Then:
$$f(x,y) = \sum_{k=0}^{+ \infty} \sum_{n=k}^{+ \infty} \binom{n}{k} \left( \frac{y}{x} \right)^k x^n = \sum_{n=0}^{+ \infty} x^n \sum_{k=0}^n \binom{n}{k} \left( \frac{y}{x} \right)^k = \sum_{n=0}^{+ \infty} x^n \left( 1+\frac{y}{x} \right)^n = 1/(1-y-x).$$
If you take the $k$-th partial derivative with respect to $y$ in $(x,0)$ and divide by $k!$, you get:
$$\sum_{n=0}^{+ \infty} \binom{n+k}{k} x^n = \frac{1}{(1-x)^{k+1}}.$$
Now, you just have to be careful about the radius of convergence of such series. Taking $|y| < 1-|x|$ should do the trick.
A: Negative Binomial Coefficients
As shown in this answer,
$$
\binom{-n}{k}=(-1)^k\binom{n+k-1}{k}
$$
Substituting $n\mapsto n+1$ then multiplying by $(-1)^k$ gives
$$
(-1)^k\binom{-n-1}{k}=\binom{n+k}{k}
$$
Thus,
$$
\begin{align}
\sum_{k=0}^\infty\binom{n+k}{k}x^k
&=\sum_{k=0}^\infty(-1)^k\binom{-n-1}{k}x^k\\
&=\sum_{k=0}^\infty\binom{-n-1}{k}(-x)^k\\[4pt]
&=(1-x)^{-n-1}\\[7pt]
&=\frac1{(1-x)^{n+1}}
\end{align}
$$

Differentiating A Geometric Series
Start with
$$
\frac1{1-x}=\sum_{k=0}^\infty x^k
$$
and take $n$ derivatives
$$
\begin{align}
\frac{n!}{(1-x)^{n+1}}
&=\sum_{k=0}^\infty k(k-1)\cdots(k-n+1)x^{k-n}\\
&=\sum_{k=n}^\infty k(k-1)\cdots(k-n+1)x^{k-n}\\
&=\sum_{k=0}^\infty(k+n)(k+n-1)\cdots(k+1)x^k\\
&=\sum_{k=0}^\infty\frac{(k+n)!}{k!}x^k
\end{align}
$$
Finally, divide by $n!$ to get
$$
\frac1{(1-x)^{n+1}}=\sum_{k=0}^\infty\binom{k+n}{k}x^k
$$
A: One has: $$\frac{1}{(1-x)^{k+1}}=\prod_{i=1}^{k+1}\frac{1}{1-x}=\prod_{i=1}^{k+1}\sum_{n=0}^{+\infty}x^n.$$
Now, let define: $$\forall n\in\mathbb{N},a_n:=\#\left\{(k_i)_{i\in\{1,\ldots,k+1\}}\in\mathbb{N}^{k+1}\textrm{ s.t. }\sum_{i=1}^{k+1}k_i=n.\right\}.$$
Hence, one has: $$\frac{1}{(1-x)^{k+1}}=\sum_{n=0}^{+\infty}a_nx^n.$$
Your problem is reduced to a combinatorial problem: for every $n\in\mathbb{N}$, you need to find $a_n$, you will find: $$a_n={n+k\choose k}.$$
A: 
A few examples suggest that the answer is $~\dfrac{1}{(1-x)^{k+1}}~,~$ but I don't see how to show this.

Of course you don't ! And do you know why ? Because of the n in $\displaystyle{\color{red}n+k\choose k}.~$ That cute little 
culprit throws a wrench into all our efforts. After all, even if we were to expand $~\dfrac1{(1-x)^{k+1}}~$ 
into its binomial series, we'd still get a constant term $($in this case, $-k-1)$ there. Instead, we 
have $n+k,~$ where n is variable. Unless we could show that $\displaystyle{n+k\choose k}=(-1)^n{-k-1\choose n}.~$ 
To do that, just rewrite $~\displaystyle{-k-1\choose n}~$ explicitly as 

$$\dfrac{\Big[(-k-1)-0\Big]\Big[(-k-1)-1\Big]\cdots\Big[(-k-1)-(n-1)\Big]}{1\cdot2\cdot3\cdots n},$$ 

and factor $-1$ out of each term in the numerator, then compare the expression with that of 

$${n+k\choose k}~=~\dfrac{\Big[(n+k)-0\Big]\Big[(n+k)-1\Big]\cdots\Big[(n+k)-(k-1)\Big]}{1\cdot2\cdot3\cdots k}~.$$ 

Depending on which of the two numbers k and n is greater, one of the two expressions will  suffer cancellations in both its numerator and denominator.
