Characteristic polynomial of a nilpotent matrix

If the overlying field is the complex numbers (see Listing's and Mark's comment):

If $A^k=0$ and $\lambda$ is an eigenvalue of $A$ with eigenvector $\bf x$:

$$\eqalign{ A {\bf x}=\lambda {\bf x} &\Rightarrow A^2 {\bf x}= \lambda^2 {\bf x} \cr &\Rightarrow A^3 {\bf x}=\lambda^3 {\bf x}\cr &\ {\vdots} \cr &\Rightarrow 0=\lambda^{k } {\bf x} \cr & \Rightarrow \lambda=0} \ \ \ \raise6pt{\left. {\vphantom{\matrix{1\cr1\cr1\cr1\cr1\cr}}}\right\}} \raise6pt{\scriptstyle{(k-1)-\text{times}}} $$

So 0 is the only eigenvalue of $A$. The characteristic polynomial of $A$ is then $x^n$.

The minimal polynomial is of the form $x^k$ for some $k$, because the matrix is nilpotent. Since the minimal polynomial is divisible by all the irreducible polynomials which divide the characteristic polynomial, we see that in fact the only irreducible polynomial which divides the latter is $x$. Thus $\chi_A(x)=x^n$.