Sums involving binomial coefficients in a finite field Consider the field $\mathbb{F}_q$ where $q=p^k$ for some prime $p$. I have some identities related to binomial coefficients over such a field, which I wish to prove. So, can someone tell me a source where I could read up on these?
An example of the identities that I am looking out for is the following. For some $a$, such that $0 \leq a < q$,
$${q(q-1)\choose(q-a)(q-1)} + {(q-1)(q-1)\choose(q-a)(q-1)} + {(q-2)(q-1)\choose(q-a)(q-1)} + . . . + {(q-a)(q-1)\choose(q-a)(q-1)} = 1$$
in $\mathbb{F}_q$. I have some more similar identities that I wish to prove and I would like it if someone could give me a hint / a strategy / a reference for the same. (Please comment if you wish to see more context)

I found the solution to the above problem. A general term of the above sum looks like ${(q-t)(q-1) \choose (q-a)(q-1)}$ which is the coefficient of $x^{(q-a)(q-1)}$ in $(1+x)^{(q-t)(q-1)}$. Hence, the sum is the coefficient of $x^{(q-a)(q-1)}$ in $(1+x)^{q(q-1)}+(1+x)^{(q-1)^2}+...+(1+x)^{(q-a)(q-1)}$. We can continue this sum up till 1 as none of those terms will alter any coefficient of $x^{(q-a)(q-1)}$. So, we are looking for the coefficient of $x^{(q-a)(q-1)}$ in:
$$(1+x)^{q(q-1)}+(1+x)^{(q-1)^2}+...+(1+x)^{(q-a)(q-1)} + (1+x)^{(q-a-1)(q-1)} + . . . + 1$$
$$=\frac{(1+x)^{(q-1)(q+1)}-1}{(1+x)^{q-1}-1}$$
$$=\frac{\frac{(1+x)^{q^2}}{1+x}-1}{\frac{(1+x)^q}{1+x}-1}$$
$$=\frac{\frac{1+x^{q^2}}{1+x}-1}{\frac{1+x^q}{1+x}-1}$$
$$=\frac{x^{q^2}-x}{x^q-x}$$
$$=\frac{x^{q^2-1}-1}{x^{q-1}-1}$$
$$=1+x^{q-1}+x^{2(q-1)}+...+x^{q(q-1)}$$
And hence, the required coefficient is $1$.
