TITLE: Proving that ${x +y+n- 1 \choose n}= \sum_{k=0}^n{x+n-k-1 \choose n-k}{y+k-1 \choose k} $
QUESTION [10 upvotes]: How can I prove that $${x +y+n- 1 \choose n}= \sum_{k=0}^n{x+n-k-1 \choose n-k}{y+k-1 \choose k} $$
I tried the following:
We use the falling factorial power: 
$$y^{\underline k}=\underbrace{y(y-1)(y-2)\ldots(y-k+1)}_{k\text{ factors}},$$
so that $\binom{y}k=\frac{y^{\underline k}}{k!} .$ 
Then
$${x +y+n- 1 \choose n} = \frac{(x +y+n- 1)!}{n! ((x +y+n- 1) - n)!} = 
\frac{1}{n!}. (x +y+n \color{#f00}{-1})^{\underline n} $$
And
$$ {x+n-k-1 \choose n-k}{y+k-1 \choose k}$$
$$\frac{1}{(n-k)!}.(x+n-k-1)^{\underline{n-k}}.\frac{1}{k!}.(y+k-1)^{\underline{k}}$$
$$\frac{1}{k!.(n-k)!}.(x+n-k-1)^{\underline{n-k}}.(y+k-1)^{\underline{k}}$$
$$\sum_{k=0}^n{n \choose k}(x+n-k-1)^{\underline{n-k}}.(y+k-1)^{\underline{k}}$$
According to the Binomial-coefficients:
$$ ((x+n-k-1) + (y+k-1))^{\underline{n}}$$
$$ (x+y+n\color{#f00}{- 2})^{\underline{n}}$$
What is wrong ? und How can I continue?  :/

REPLY [3 votes]: Hint: The binomial formula with the Cauchy product
  \begin{align*}
(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^ky^{n-k}
\end{align*}
  does not use falling factorials $x^{\underline{k}}$ resp. $y^{\underline{n-k}}$.

Here   is a step by step answer similar to that by @MarkoRiedel.  It's convenient to use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write e.g.
\begin{align*}
\binom{n}{k}=[z^k](1+z)^n
\end{align*}

We obtain 
\begin{align*}
\sum_{k=0}^{n}&\binom{x-1+n-k}{n-k}\binom{y-1+k}{k}\\
&=\sum_{k=0}^{\infty}\binom{x-1+n-k}{n-k}\binom{-y}{k}(-1)^k\tag{1}\\
&=\sum_{k=0}^\infty [t^{n-k}](1+t)^{x-1+n-k}[z^k](1+z)^{-y}(-1)^k\tag{2}\\
&=[t^n](1+t)^{x-1+n}\sum_{k=0}^\infty(-1)^kt^k(1+t)^{-k}[z^k](1+z)^{-y}\tag{3}\\
&=[t^n](1+t)^{x-1+n}\sum_{k=0}^\infty\left(-\frac{t}{1+t}\right)^k[z^k](1+z)^{-y}\\
&=[t^n](1+t)^{x-1+n}\left(1-\frac{t}{1+t}\right)^{-y}\tag{4}\\
&=[t^n](1+t)^{x+y-1+n}\tag{5}\\
&=\binom{x+y-1+n}{n}
\end{align*}
  and the claim follows.

Comment:

In (1) we use  the binomial identity $\binom{-p}{q}(-1)^q=\binom{p+q-1}{q}$ and we extend the upper limit of the series to $\infty$ without changing anything since we are adding zeros only.
In (2) we apply the coefficient of operator twice.
In (3) we do some rearrangements by using the linearity of the coefficient of operator  and  we also  use  the  rule
\begin{align*}
[z^{p-q}]A(z)=[z^p]z^{q}A(z)
\end{align*}
In (4) we apply the substitution rule
\begin{align*}
A(t)=\sum_{k=0}^\infty a_kt^k=\sum_{k=0}^\infty t^k[z^k]A(z)\\
\end{align*}
with $z=-\frac{t}{1+t}$.
In (5) we do some simplifications.
In (6) we select the coefficient from $t^n$.