Is there a more concise expression of this product?

Question

In a longer computation, I have stumbled upon the following product, where $k,r \in \mathbb{N}_0$ are fixed numbers: $$\prod_{0 < i_0<i_1<\dots<i_r\leq k} (i_r-i_{r-1})(i_{r-1}-i_{r-2})\cdots(i_1-i_0) \ .$$ This product seems somehow familiar and I am quite sure that it might be expressed in a more concise way in terms of factorials, but somehow I fail to find the right expression...

Answer 1

that product is clearly equal to:

$$\prod\limits_{l=r+1}^k\prod\limits_{a_1+a_2+\dots +a_{r+1}=l}a_1a_2\dots a_{r+1}$$ with $a_i>0$

We first find a closed form for the second factor.

How many times does each factor $a$ between $1$ and $l-r$ appear?

If we only count when it appears in position $1$ we get:

$\binom{l-a-1}{r-1}$, since the remaining $r$ terms must add up to $k-a$ (so it is a problem of counting weak compositions).

Therefore your product is:

$$\prod\limits_{l=r+1}^k\prod_{j=1}^{l-r}j^{(r+1)\binom{l-j-1}{r-1}}=\prod\limits_{l=r+1}^k(\prod_{j=1}^{l-r}j^{\binom{l-j-1}{r-1}})^{r+1}$$

We now use:

$$j^{\binom{r-1}{r-1}}\times j^{\binom{r-1+1}{r-1}}\times\dots\times j^{\binom{k-j-1}{r-1}}=j^{\binom{r-1}{r-1}+\dots \binom{k-j-1}{r-1}}$$

Finally, we have $\binom{r-1}{r-1}+\dots \binom{k-j-1}{r-1}=\binom{k-j}{r}$ due to the hockey stick equality.

So the final form is:

$$(\prod\limits_{j=1}^{k-r}j^{\binom{k-j}{r}})^{r+1}$$