Prove $\binom{N+K-1}{K}=\sum_{i=1}^{N-1} (-1)^{i+1}\binom{N}{i}\binom{N-i+K-1}{K}$ using polynomial I find the identity, $$\binom{N+K-1}{K}=\sum_{i=1}^{N-1} (-1)^{i+1}\binom{N}{i}\binom{N-i+K-1}{K}$$
by counting the number of ways to distribute $K$ candies to $N$ people ($N>K)$.
For right-hand eq., I count the number of ways to distribute candies when there is at least $i$ people who cannot get candies.
Can we prove this identity using polynomials?
For left-hand eq., it corresponds to the coefficient of $x^K$ of $\frac{1}{(1-x)^N}$. Can we get the right-hand eq. by manipulating this polynomial?
 A: We can re-write the identity as
$$\sum_{q=0}^{N-1} (-1)^{q+1} {N\choose q} {N-q+K-1\choose K} = 0.$$
Equivalently (divide by $-1$),
$$\sum_{q=0}^{N-1} (-1)^q {N\choose q}
{N-q+K-1\choose N-1-q}
\\ = \sum_{q=0}^{N-1} (-1)^q {N\choose q}
[z^{N-1-q}] (1+z)^{N-q+K-1}
\\ = [z^{N-1}] (1+z)^{N+K-1}
\sum_{q=0}^{N-1} (-1)^q {N\choose q}
z^q (1+z)^{-q}.$$
Now we may add in the term for $q=N$ because it does not contribute to
the coefficient extractor:
$$[z^{N-1}] (1+z)^{N+K-1}
\sum_{q=0}^{N} (-1)^q {N\choose q}
z^q (1+z)^{-q}
\\ = [z^{N-1}] (1+z)^{N+K-1}
\left(1-\frac{z}{1+z}\right)^N
\\ = [z^{N-1}] (1+z)^{K-1}.$$
This  is zero  by  inspection  when $N\gt  K$  as  claimed. We  obtain
${K-1\choose N-1}$ when $N\le K.$
 Remark. If  an inverse binomial is insisted upon  we may start
from
$$\sum_{q=0}^{N-1} (-1)^q {N\choose q} {N-q+K-1\choose K}
\\ = \sum_{q=0}^{N-1} (-1)^q {N\choose q}
[z^{N-1-q}] \frac{1}{(1-z)^{K+1}}
\\ = [z^{N-1}] \frac{1}{(1-z)^{K+1}}
 \sum_{q=0}^{N-1} (-1)^q {N\choose q} z^q.$$
Same action of the coefficient extractor as before:
$$[z^{N-1}] \frac{1}{(1-z)^{K+1}}
\sum_{q=0}^{N} (-1)^q {N\choose q} z^q
= [z^{N-1}] \frac{1}{(1-z)^{K+1}} (1-z)^N
\\ = [z^{N-1}] \frac{1}{(1-z)^{K-N+1}}.$$
Now when $N\gt K$ this becomes $[z^{N-1}] (1-z)^{N-K-1}$ which is zero
by  inspection.   With  $N\le  K$  we  find  ${N-1+K-N\choose  K-N}  =
{K-1\choose K-N}  = {K-1\choose N-1}$  in agreement with  the previous
result. We suppose in both versions that $N,K\ge 1.$
