But Hollywood, 548 is NOT a powerful number; as you said 548 = 137*2^2, and a number is powerful if every prime in its prime factorisation appears to the power of 2 or a higher power, and 137 appears only to the power of 1 in the prime factorisation of 548.That invalidates your counterexample.
If n is a perfect square, then log(N(n)) = log(1) = 0, so that power(n) is undefined. Otherwise, if n is powerful, it would seem that each prime in the prime factorization of n would add at least 3 times the value to the numerator as the denominator, so that power(n) is greater than or equal to 3.
I was thinking that the condition was "n is not squarefree" instead of "n is powerful". Sloppy reading, I guess.....
So for some small non-squarefree (or "squareful") numbers (like for 12, power(n)=log(12)/log(3)=2.26...), power(n) is greater than two. You need to have a big prime factor to the first power to get power(n) down below 2. Actually, 20 = 5*2^2 is the first squareful number for which power(n) is less than 2, so I didn't need to go all the way up to 548.