That looks to me like a basic "combinatorics" calculation. Since U has dimension "k" over F, each vector in U can be written as a ordered k-tuple of elements of F. F contains "q" objects so the number of ways to do that is q(q)...(q) k times:
That looks to me like a basic "combinatorics" calculation. Since U has dimension "k" over F, each vector in U can be written as a ordered k-tuple of elements of F. F contains "q" objects so the number of ways to do that is q(q)...(q) k times: