Suppose we have a prime $p$ and consider $\mathbb{F}_q$ where $q=p^s$ for some $s$. Fix a positive integer $m \geq 2$ and let $t \leq m-1$. Let $r$ be a positive integer such that $0 \leq r \leq q^t-1$. Then, consider the sum for any $j$ such that $0 \leq j \leq r$:
$$\sum\limits_{i=0}^{\frac{q^m - q^t - 2q + 2}{q-1}}{(q-1)(\frac{q^m-q}{q-1}-i) \choose j}{(q-1)(\frac{q^t+q-2}{q-1}+i) \choose r-j}$$
(where the binomial coefficients are assumed to be $0$ if they do not make sense) I wish to show that this sum is independent of $j$. This (convoluted!) identity comes from my study of invariant polynomials in $2$ variables over $\mathbb{F}_q$. (For the purpose of this question, replacing $\mathbb{F}_q$ by $\mathbb{F}_p$ doesn't make a difference) For more context, please comment.
I think that the best way to attack this would be to use Lucas' Theorem to simplify the binomial coefficients and then, think of this sum as the coefficient of a particular monomial in a polynomial expression. This works successfully in the special case $m=2$ and $t=1$, but I haven't been able to extend the solution to the general case. Any help/hints will be appreciated.