TITLE: Showing $\binom{2d-1}{n}-(n+1){d-1\choose n}=\sum_{i=1}^n (-1)^{n-i}{n+1\choose i+1}{id-d+n\choose n}$ for $n\geq1$ and $d\geq n+1$
QUESTION [6 upvotes]: Recently, when I was trying to compute Hodge numbers of hypersurfaces in toric varieties, I discovered the following combinatorial identity:  

For every positive integers $n\geq 1$ and $d\geq n+1$ the following is an identity
   $$
   \binom{2d-1}{n}-(n+1){d-1\choose n}=\sum_{i=1}^n (-1)^{n-i}{n+1\choose i+1}{id-d+n\choose n}
 $$

I tried to prove it by induction on $n$; for $n=1$ the identity becomes trivial, but the inductive step seems to be very hard (at least for me!).
For $d=n+1$ the identity can be proved by using arguments of V. Batyrev involving polar duality of reflexive polytopes. But for greater $d$ I have not any idea...
Someone can help me?
Thank you a lot!

REPLY [2 votes]: First note that as an order $n+1$ repeated difference of a degree $n$ polynomial,
$$
\overbrace{\sum_{i=-1}^n(-1)^{n-i}\binom{n+1}{i+1}}^\text{order $n+1$ difference}\overbrace{\binom{id-d+n}{n}\vphantom{\sum_{i=-1}^n}}^\text{degree $n$ polynomial}=0\tag1
$$
The difference between $(1)$ and the sum in the question are the $i=-1$ and $i=0$ terms. Thus,
$$
\begin{align}
\sum_{i=1}^n(-1)^{n-i}\binom{n+1}{i+1}\binom{id-d+n}{n}
&=\textstyle\overbrace{\color{#C00}{(-1)^n\binom{n+1}{0}\binom{-2d+n}{n}}}^{\text{negative of $i=-1$ term of $(1)$}}-\overbrace{\color{#090}{(-1)^n\binom{n+1}{1}\binom{-d+n}{n}}}^{\text{$i=0$ term of $(1)$}}\\[3pt]
&=\color{#C00}{\binom{2d-1}{n}}-\color{#090}{(n+1)\binom{d-1}{n}}\tag2
\end{align}
$$
where in the last step of $(2)$, we evaluated the negative binomial coefficients.