Binomial Coefficient Identity Involving Summation Prove that 
$$\sum_{j=0}^n (-1)^j \binom{n+j-1}{j}\binom{N+n}{n-j} = \binom{N}{n} $$
I tried to prove this via binomial expansions of $(1-x)^N (1+x)^{-m}$, and equating the coefficients of $x$, although I am unsure about my approach. Could someone please post a solution/
 A: From expanding out ${-n\choose j}$ we see that
$$ (-1)^j{n+j-1\choose j}={-n\choose j}$$
and
$$ \sum_{j=0}^n{-n\choose j}{N+n\choose n-j}={N\choose n} $$
by looking at the coefficient of $x^n$ in $(1+x)^N=(1+x)^{-n}(1+x)^{N+n}$.
A: This  one  is  extremely  straightforward using  binomial  coefficient
integral   representation.   For   future  reference,   this   is  the
calculation.
Introduce
$${N+n\choose n-j}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-j+1}} (1+z)^{N+n} \; dz.$$
Observe  that  this  vanishes when  $j\gt  n$  so  we may  extend  the
summation to infinity, getting
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{N+n}
\sum_{j\ge 0} (-1)^j {n-1+j\choose j} z^j \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{N+n}
\sum_{j\ge 0} (-1)^j {n-1+j\choose n-1} z^j \; dz.$$
This simplifies (Newton binomial) to
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{N+n}
\frac{1}{(1+z)^n}\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{N}
\; dz
\\ = {N\choose n}.$$
