you should post questions like this in the algebra section if you want to get a fater response.
anyway, suppose first that every
-dimesnional 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 eigenvector of
. then there exists and
such that
but then 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.