I tried doing this, but kept failing to prove. I know how to prove that the language is nonregular when n0(w) = n1(w). The following is the proof for n0(w) = n1(w) using pumping lemma:
Assume L is regular. Then, by the Pumping Lemma, there is a natural number m such that any w ∈ L with |w| ≥ m can be factored as w = xyz with |xy| ≤ m and |y| > 0, and xyi z ∈ L, for i = 0, 1... Pick w = 0^m 1^m. Then, 0^m 1^m = xyz, where y = 0k, for k > 0. By the Pumping Lemma, xz ∈ L. But, n0(xz) does not equal n1(xz). The assumption that L is regular thus is false. Hence Proved
However, I don't know how to prove it for when number of 0 is not equal to number of 1.