[Math] Prove that the characteristic polynomial of a nilpotent matrix is $x^n$

Question

How can I prove that the char.pol. of a nilpotent matrix is of the form $x^k$?
I'm trying to do it by contradiction but assuming that $p_{xA}=a_0+a_1x+\dots+a_mx^m+\dots+a_nx^n$ seems not giving any contradiction.
I've proved that the eigenvalues of A has to be 0, which led to $det(A)=0$.

Answer 1

If $A^k = 0$, then $(\lambda I - A)(\lambda^{k - 1}I + \lambda^{k - 2}A + \cdots + A^{k - 1}) = \lambda^kI$. Taking determinants, we get that the characteristic polynomial of $A$ divides $det(\lambda^kI) = \lambda^{nk}$. Since the characteristic polynomial of $A$ is a monic polynomial of degree $n$, it follows that it must be $\lambda^n$.