How to prove that the nullities of powers of a nilpotent matrix are not equal? 
Let $A$ be a nilpotent matrix of order $n$ . Prove that if $A^m \neq 0$, where $m \in \Bbb Z^{+}$, then the nullity of $A^m$ is smaller than the nullity of $A^{m+1}$.

I understand the following are the thought process:
Prove that the nullspace of $A^m$ is contained in the nullspace of $A^{m+1}$,
--> Do we use linear independence?
then prove that they are not equal.
--> Which theory can be used?
Since $A^m \neq 0$, there exists $u \in \Bbb R^n$ such that $A^m u \neq 0$. Then select a proper vector from
$$A^m u, A^{m+1}u, A^{m+2}u,\dots$$
which is in the nullspace of $A^{m+1}$ but is not in the nullspace of $A^m$.
--> I don't really understand how to do this?
 A: By definition, the nullity of a matrix is the dimension of its kernel (AKA nullspace).  We see that $\ker(A^{m + 1}) \subseteq \ker(A^m)$ for any matrix $A$ by noting that if a vector $x$ is such that $A^mx = 0$, then we must have
$$
A^{m+1}x = A(A^mx) = A(0) = 0
$$
So, if $x$ is in the nullspace of $A^m$, then it must also be in the nullspace of $A^{m+1}$.
Now, we have to use the nilpotent propterty to prove that $\ker(A^{m+1}) \neq \ker (A^m)$.  From there, we can use the fact that $\ker(A^{m+1}) \subsetneq \ker (A^m)$ to conclude that $\dim \ker (A^{m+1}) \lneq \dim \ker(A^m)$.  That is, we will be able to conclude that the nullity of $A^{m+1}$ is less than that of $A^m$.

Now, let's follow their hint: we know that $A^m \neq 0$, so there is a vector $u$ such that $A^m u  \neq 0$.  However, $A^nu = 0$, so we know that we must be able to find a $k$ with $1 \leq  k \leq n-k$ such that $A^{m+k}u = 0$, but $A^{m+k-1}u \neq 0$.  
Look the list of vectors $A^mu,A^{m+1}u,A^{m+2}u,\dots$ and think about why the above is true. It should be "obvious".
Once we have established that this is the case, I claim that $A^{k-1}u$ is in the nullspace of $A^{m+1}$ but not in the null space of $A^m$ (by definition, $A^0u = u$).  To see this, note that
$$
A^{m}(A^{k-1}u) = A^{m + k - 1}u \neq 0\\
A^{m+1}(A^{k-1}u) = A^{m+k}u = 0
$$
Now, we've shown that $\ker(A^{m+1}) \neq \ker (A^m)$.  But, with what I said earlier, the proof is complete.
