No, because you don't compute n^n by multiplying n by n, n times. Your also don't compute a x b by summing b, a times, do you? You compute n^n the same way you

No, because you don't compute n^n by multiplying n by n, n times. Your also don't compute a x b by summing b, a times, do you? You compute n^n the same way you compute a^n which is to say that a^n = e^(n x log(a))