I found a "surprising" result involving the $k$-th Difference of product of multiple arithmetic progression powers but am not sure how to prove it. Hope somebody here can help me to prove.
Given $n$ arithmetic progressions with initial terms $a_{i}$ and the common differences of successive members $d_i$ and powers $p_i$ $(i=1, \dots, n)$, we can form a product as following:
$a_{1}^{p_1}\times \dots \times a_{n}^{p_n}, (a_{1}+d_1)^{p_1}\times \dots \times (a_{n}+d_n)^{p_n}, (a_{1}+2d_1)^{p_1}\times \dots \times (a_{n}+2d_n)^{p_n},\dots, (a_{1}+j\cdot d_1)^{p_1}\times \dots \times (a_{n}+j\cdot d_n)^{p_n},\dots$
Now one can form $1$-difference(the difference between this term and the previous term), $2$-difference (the $1$-difference of $1$-difference), ..., $k$-difference on such product. What I found it that: after $p=\sum_{i=1}^np_{i}$-difference, we will get back a series with constant terms which are all $p!\prod_{i=1}^nd_{i}^{p_i}$.
Directly writing out every term and do the difference up to $p$-th is very cumbersome. There must be some trick to do this, but I am not seeing how.
Thank you for your help.