If a Matrix Has Only Zero as an Eigen-Value Then It Is Nilpotent

We have to assume that we are considering complex matrices. Over the reals the assertion is not true, as the example $$ \begin{bmatrix}0&0&0\\0&0&1\\0&-1&0 \end{bmatrix} $$ shows.

Over $\mathbb C$, one can do the Schur decomposition, where $A=VTV^*$, with $V$ unitary and $T$ upper triangular. Since the diagonal of $T$ has to contain the eigenvalues of $A$, it has be zero. And it is an easy exercise that if $T$ is an $n\times n$ upper triangular with diagonal zero, then $T^n=0$. So $A^n=(VTV^*)^n=VT^nV^*=0$, and $A$ is nilpotent.

Hint: $A = P^{-1}DP$ where $D$ is upper triangular. What are the diagonal entries of $D$? If $A$ is $n\times n$, what is $A^n$?

Hint: Look at the characteristic polynomial, then use Cayley-Hamilton.