How to prove $\sum\limits_{i=0}^{k}(-1)^i\binom{n}{k-i}\binom{n+i-1}{i}=0$ I saw a combinatorial identity when i study linear-algebra, But the author  didn't explain how to get it.

$\displaystyle \sum_{i=0}^{k}(-1)^i\binom{n}{k-i}\binom{n+i-1}{i}=0$

Who can help me to prove it? Thanks!

You probably want to require $k > 0$, as otherwise your claim is $1 = 0$.
Renaming $k$, $n$ and $i$ as $n$, $x$ and $k$, you get a particular case (the case $x=y$) of the following identity:
$\sum\limits_{k=0}^n \left(-1\right)^k \dbinom{x}{n-k} \dbinom{k+y-1}{k} = \dbinom{x-y}{n}$.
This is, e.g., Proposition 3.32 (d) in my Notes on the combinatorial
fundamentals of algebra. (In case of changing numbering, search for " some sample applications of Theorem", or check out the frozen version of 10 January 2019, in which the numbering surely has not shifted.) The main trick to the proof is realizing that $\left(-1\right)^k \dbinom{k+y-1}{k}$ can be rewritten as $\dbinom{-y}{k}$, after which you are left with a particular case of the Vandermonde convolution identity.
