Equality like Pascal triangle I have noticed the following is true. 
Let's denote the equation below as (1)
$$\sum_{k=1}^{d+1}(-1)^{d+1-k}\frac{1}{(d+1)(k-1)!(d+1-k)!}\prod_{i=1}^{d+1}\Big(\frac{q}{h}+(k-i)\Big)\prod_{j=k}^{d}\Big(1-jh\Big)\prod_{\ell=d-k+2}^{d}\Big(1+\ell h\Big).$$
Let's denote the equation below as (2) 
$$\sum_{k=1}^{d}(-1)^{d-k}\frac{1}{(d)(k-1)!(d-k)!}\prod_{i=1}^{d}\Big(\frac{q}{h}+(k-i)\Big)\prod_{j=1}^{d}\Big(1+(k-j)h\Big).$$
If I replace $q\, \text{by}\, \frac{1}{x-1}$ in eq(1)
and $q\text{by}\, 1/x$ and $h\,\text{by}\, \frac{(x-1)h}{x}$ in eq (2) we have tested for few values of  $d$
$(x-1)^{d}\, eq(1) = x^d \,\text{eq (2)}$
Can I have a proof of this fact? Any help or why is true? 
Notice that eq (1) has $d+1$ terms and eq(2) has $d$ terms. So its seems like pascal triangle any combinatorial viewpoint? 
 A: This is a partial answer; it gives a computer-aided proof (yes, the identity holds true).
Some cleanup first. We have
$$\prod_{j=k}^{d}(1-jh)\prod_{\ell=d-k+2}^{d}(1+\ell h)=\frac{\prod_{\ell=-d}^{d}(1+\ell h)}{\prod_{j=1}^{d+1}\big(1+(j-k)h\big)},$$
so that $(1)$ equals
$$\frac{\prod_{\ell=-d}^{d}(1+\ell h)}{(-h)^{d+1}(d+1)!}\sum_{k=0}^{d}(-1)^{d-k}\binom{d}{k}\prod_{j=0}^{d}\frac{(q/h)+k-j}{(-1/h)+k-j}.$$
Now, if we do all the suggested substitutions, denote $a=1/\big((x-1)h\big)$ and $b=-1/h$, and retain the (common) notation $\binom{x}{n}=\frac{x(x-1)\ldots(x-n+1)}{n!}$ for real $x$, we see the following needed to be proven.

Let $$F_d(a,b)=\sum_{k=0}^{d-1}(-1)^k\binom{d-1}{k}\binom{a+k}{d}\binom{b+k}{d},\\G_d(a,b)=\sum_{k=0}^{d-1}(-1)^k\binom{d-1}{k}\binom{a+k}{d}\Big/\binom{b+k}{d}.$$ Then $$F_d(a,a-b)=(-1)^{d+1}\binom{2d+1}{d}\binom{b+d}{2d+1}G_{d+1}(a,b).\tag{*}$$

I don't see an "elegant" proof yet. But the book linked above suggests methods to find recurrences satisfied by $F_d$ and $G_d$, via "telescoping" recurrences satisfied by the summands. Specifically, in our case we find
$$d(a-b-d-1)(a-b+d+1)F_d(a,b)\\+(d+1)(2d+3)(a+b)F_{d+1}(a,b)\\+(d+1)(d+2)(d+3)F_{d+2}(a,b)=0$$
(using a computer) and
$$(d-1)(d+1)G_d(a,b)\\+(2d+1)(b-2a)G_{d+1}(a,b)\\-(b-d-1)(b+d+1)G_{d+2}(a,b)=0.$$
Now it's easy to verify that $(\text{*})$ holds for $d\in\{1,2\}$, and that both sides of it satisfy the same recurrence.
