Yes, that certainly does help!

I presume that in doing that you also showed that if v is an eigenvector of A corresponding to eigenvalue $\displaystyle \lambda$ then v is also an eigenvector of $\displaystyle A^m$ corresponding to eigenvalue $\displaystyle \lambda^m$.

In particular, if $\displaystyle \lambda$ an eigenvalue of A, then $\displaystyle \lambda^k$ is an eigenvalue of $\displaystyle A^k$- that is, for some non-zero vector v, $\displaystyle A^k v= \lambda^k v$. But $\displaystyle A^kv= 0$ for any vector so we have $\displaystyle \lambda^k v= 0$, with v non-zero.