$ \sum_{k=0}^n (-1)^{k} {{m+1}\choose k }{{m+n-k}\choose m }= 0$ if $n>0$ I need to show that  $ \sum_{k=0}^n (-1)^{k} {{m+1}\choose k }{{m+n-k}\choose m }= 0 $ if $n>0$. Here $m$ is a non negative integer.
I am thinking induction, but do I apply it on $m$ or $n$? I have tried both. Please help.
 A: Starting from
$$\sum_{k=0}^n (-1)^k {m+1\choose k} {m+n-k\choose m}
= \sum_{k=0}^n (-1)^k {m+1\choose k} {m+n-k\choose n-k}$$
we find
$$[z^n] (1+z)^{m+n}
\sum_{k=0}^n (-1)^k {m+1\choose k} z^k (1+z)^{-k}.$$
Now we may extend  $k$ beyond $n$ because of the  factor $z^k$ and the
coefficient extractor $[z^n]$ in front:
$$[z^n] (1+z)^{m+n}
\sum_{k\ge 0} (-1)^k {m+1\choose k} z^k (1+z)^{-k}
\\ = [z^n] (1+z)^{m+n} \left(1-\frac{z}{1+z}\right)^{m+1}
= [z^n] (1+z)^{n-1} = 0.$$
A: Also, we can write it as 
$$
\eqalign{
  & \sum\limits_{0\, \le \,k\, \le \,n} {\left( { - 1} \right)^{\,k} \left( \matrix{
  m + 1 \cr 
  k \cr}  \right)\left( \matrix{
  m + n - k \cr 
  m \cr}  \right)} \quad \left| {\,0 \le m} \right.\quad  =   \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( { - 1} \right)^{\,k} \left( \matrix{
  m + 1 \cr 
  k \cr}  \right)\left( \matrix{
  m + n - k \cr 
  n - k \cr}  \right)}  =   \quad \quad (1)\cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( \matrix{
  k - m - 2 \cr 
  k \cr}  \right)\left( \matrix{
  m + n - k \cr 
  n - k \cr}  \right)}  = \quad \quad (2)  \cr 
  &  = \left( \matrix{
  n - 1 \cr 
  n \cr}  \right)\quad \quad \quad \quad (3) \cr} 
$$
where:
 -(1) we can apply "symmetry" since $m$ and $n-k$ are non negative;
 -(2) upper negation, and we can omit the bounds because implicit in the binomials;
 -(3) "double convolution".
