Matrix with given characteristic polynomial is similar to its square I have a test on Wednesday and I've been dealing with this problem for quite a while. I would greatly appreciate any help or advice on how to solve it.

Given a matrix $A$ in $\mathbb C^{n\times n}$ (space of $n$ by $n$ complex matrices), whose characteristic polynomial is $(x - 1)^n$. Prove that $A^2$ is similar to $A$.

I know that given the characteristic polynomial of $A$ posted above, it follows that
its Jordan form has ones all along the diagonal and zeroes above (I use the lower triangular convention for Jordan form). So $A$'s determinant is equal to one and $A$ is invertible (same thing for $A^2$).
I just can't get past this point.
Thank you in advance for your time.

Hint
Use the Jordan-Weyr theorem See this paper in French
$$A\sim B\iff \mathrm{rank}(A-\lambda I)^k=\mathrm{rank}(B-\lambda I)^k,\quad \forall k\in \mathbb N_{\geq0}$$
and 
$$\text{if}\ B\ \text{is invertible matrix then}\ \mathrm{rank}(AB)=\mathrm{rank}(A)$$
