You've almost got it. If, by contradiction, A is diagonalizable and all its eigenvalues are 0, then what's A's corresponding diagonal matrix? Do you see why that is a contradiction ( )?
Let such that for some positive integer . If is a nonzero matrix, show that is not diagonalizable.
I proved that if for some positive integer , then the only eigenvalue of is . I think it can be proven by contradiction. But I'm still struggle.