In the above case, there should be a factor of 'n' for the number of possible initial states.(Starting state can be chosen from n states in the dfa)
Now for nfa case:
each alphabet at each state has 2^n choices to chose from(total number of subsets of states possible).So total 2^mn at each state.Thus a total of 2^(n*n*m) combining all states together.
n choices for initial state
and 2^n choices for final states as well.
This is what I think should happen!