My question is to show the below equality
$$\sum_{j=0}^{d}(-1)^{j}\binom{d}{j}\binom{d-j}{i}=0$$ when $d>i \geq 0$ for any integers $d,i$.
This inequality is came from Stanley's note. Given an $f$-vector of a simplicial complex $\Delta$ with dimension $d-1=\dim \Delta$, $(f_0,\cdots,f_{d-1},0,\cdots)$, define a function $\mathbb{N} \to \mathbb{N}$ by sending $m$ to
$$H(\Delta,m):=1 \text{ if }m=0 \text{ or } \sum_{i=0}^{d-1}f_{i}\binom{m-1}{i} \text{ otherwise}.$$ Then, he claims that the series $$(1-\lambda)^{d}\sum_{m=0}^{\infty}H(\Delta,m)\lambda^{m}$$ as a notation $\sum_{i=0}^{\infty}h_{i}\lambda^{i}$ is indeed a polynomial with $h_{k}=0$ for all $k>d$. Since he said $h_{k}=0$ is easily seen, I try to recover it and stucked..
What I'm try to do is following; notes that when $k>d$, $$h_{k}=\sum_{j=0}^{d}(-1)^{j}\binom{d}{j}\sum_{i=0}^{d-1}f_{i}\binom{k-j-1}{i} = \sum_{i=0}^{d-1}f_{i}\sum_{j=0}^{d}(-1)^{j}\binom{d}{j}\binom{k-j-1}{i}.$$ And, indeed, by the pascal's formula and induction, it suffices to show that $$\sum_{j=0}^{d}(-1)^{j}\binom{d}{j}\binom{d-j}{i}=0$$ for all $i<d$. Also, $i=0$ is just trivial; however, I don't know how to show it when $i>0$. Do you have any hints?