Consider the polynomial:
$$\Bigg(\frac{1-t^{q+1}}{1-t}\Bigg)^{2q-1}$$
for some positive integer $q$. I wish to compute the coefficients of $t^{q^2+q+1}$ in the above polynomial. One can use the binomial theorem for a negative index maybe, but that would make the calculation very cumbersome. Is there an easier method? (If that makes it easier, we can assume that $q$ is a prime power and we are working in $\mathbb{F}_q[t]$.)
As suggested in the comments the identity $$\frac{1-t^{q+1}}{1-t}=\sum_{i=0}^qt^i=1+t+t^2+\ldots+t^q,$$ helps a lot. This shows that the coefficient of $t^{q^2+q+1}$ in the $2q-1$-th power of this expression is precisely the number of ways in which $q^2+q+1$ can be partitioned into at most $2q-1$ pieces of size at most $q$. Much has been written about these numbers, for more information see here.
Alternatively, the multinomial theorem gives us another expression for your polynomial: $$\left(\sum_{i=0}^qt^i\right)^{2q-1}=\sum_{k_0+\ldots+k_q=2q-1}\binom{2q-1}{k_0,\ldots,k_q}t^{\sum_{i=0}^qik_i}.$$ It follows that the coefficient of $t^{q^2+q+1}$ is given by $$\sum\binom{2q-1}{k_0,\ldots,k_q},$$ where the sum ranges over all $q+1$-tuples $(k_0,\ldots,k_q)$ of nonnegative integers satisfying $$\sum k_i=2q-1\qquad\text{ and }\qquad\sum ik_i=q^2+q+1.$$