anyway, suppose first that every -dimensional subalgebra of is abelian and let we only need to show that is nilpotent because then we will be done by Engel's theoem. so suppose that is an eigenvalue of and is an eigenvector corresponding to then so the subalgebra generated by would be at most -dimensional and thus, by our hypothesis, abelian (note that 1-dimensional algebras are always abelian). hence which implies so every eigenvalue of is zero and therefore is nilpotent.
conversely, suppose that is nilpotent. so for some integer now, let be a subalgebra of and suppose that is not abelian. then there exist such that for some and we may assume that so and and, if we continue this process, we will eventually get because but and and so this contradiction proves that is abelian.