6/6     If something is c*N^(M-1), is it O(N^(M-1)) or O(N^(M))?
        \_ N^(M) = N*(N^(M-1)  ==> O(N^(M-1)) is a better fit.
        \_ Yes. (wrong answer deleted -- N is not a constant)
        \_ You show a distinct lack of understanding of what O() notation
           is about.
           \_ Not really.  It's a good question.  We typically drop constants
              like '-1', but N^(M-1) == N^M / N which means that the '-1' isn't
              a constant at all. --not OP