A combinatorial identity: $\sum _{i + j = k} (-1)^i {n \choose i} {n + j - 1 \choose n - 1 } = 0 $ I proved this combinatorial identity while doing some linear algebra.
For any positive integer $k$,
$$ \sum _{i + j = k} (-1)^i {n \choose i} {n + j - 1 \choose n - 1 } = 0 $$
I was wondering what could be other ways of proving this.
EDIT: 
I thought I should mention how this identity comes up in linear algebra. For a vector space $V$ of dimension $n$, let $\Lambda(V)$ denote the exterior algebra and $ S(V) $ denote the symmetric algebra. There is a long exact sequence of vector spaces with $k + 1$ terms whose $i^{th}$ term is $$\Lambda ^ {i-1}(V) \otimes S^{k - i +1}(V)$$ One can then find the Euler characteristic of this exact sequence to get the above identity. It a question in one of the exercises in the book by Brocker-Tom Dieck on Lie groups.
 A: One can give a purely combinatorial argument. First rewrite the summation as
$$\sum_{i=0}^k(-1)^i\binom{n}i\binom{n+k-i-1}{n-1}\;.\tag{1}$$
Suppose that we have $k$ indistiguishable balls to be distributed amongst $n$ distinguishable boxes. Let $S$ be fixed set of the $n$ boxes, and let $i=|S|$. We can put one ball into each box in $S$ and then distribute the remaining $k-i$ balls in $\binom{n+k-i-1}{n-1}$ ways, so $\binom{n+k-i-1}{n-1}$ is the number of ways of distributing the $k$ balls so that each box in $S$ is non-empty. By a straightforward inclusion-exclusion argument $(1)$ is the number of ways to distribute the $k$ balls so that every box is empty, which is clearly $0$ for all $k\ge 1$.
A: The power series approach using:
$$(1-z)^{-n} =\sum_{i=0}^\infty \binom{n+i-1}{n-1}z^i$$
[You can prove this by induction on $n$, using repeated differentiation, or other methods.]
Multiply by the polynomial $(1-z)^n$, and note that, when $k>0$, the coefficient of the $z^k$ should be $0$ in the result.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{\sum _{i\ +\ j\ =\ k}\pars{-1}^{i}{n \choose i}{n + j - 1 \choose n - 1}
     = 0:\ {\large ?}}$.

\begin{align}&\color{#66f}{\large%
\sum _{i\ +\ j\ =\ k}\pars{-1}^{i}{n \choose i}{n + j - 1 \choose n - 1}}
\\[5mm]&=\sum _{j\ =\ 0}^{\infty}\ \sum _{m\ =\ 0}^{\infty}
\pars{-1}^{m}{n \choose m}{n + j - 1 \choose j}\ \overbrace{%
\oint_{\verts{z}\ =\ 1^{-}}\,\,\,\,
{1 \over z^{-m - j + k + 1}}\,\,{\dd z \over 2\pi\ic}}
^{\dsc{\delta_{m\ +\ j,k}}}
\\[5mm]&=\oint_{\verts{z}\ =\ 1^{-}}\,\,\,\,{1 \over z^{k + 1}}\
\overbrace{\bracks{\sum _{m\ =\ 0}^{\infty}{n \choose m}\pars{-z}^{m}}}
^{\dsc{\pars{1 - z}^{n}}}\ \overbrace{%
\bracks{\sum _{j\ =\ 0}^{\infty}{-n \choose j}\pars{-z}^{\,j}}}
^{\dsc{\pars{1 - z}^{-n}}}\
\,{\dd z \over 2\pi\ic}
\\[5mm]&=\oint_{\verts{z}\ =\ 1^{-}}\,\,\,\,{1 \over z^{k + 1}}
\,{\dd z \over 2\pi\ic}
=\color{#66f}{\large\left\{\begin{array}{lcrcl}
0 & \color{#000}{\mbox{if}} & k & \not= & 0
\\[1mm]
1 & \color{#000}{\mbox{if}} & k & = & 0
\end{array}\right.}
\end{align}
