$A$ is normal and nilpotent, show $A=0$

Here is a proof without using eigenvalues or diagonalization. In the below we prove the statement that "if $A^k=0$ and $k>1$ then $A^{k-1}=0$". The result then follows immediately.

1. Let $B=A^{k-1}$. Then $B$ is normal and $B^2=0$ (because $k>1$).
2. For all $x$, we have $\|B^\ast Bx\|^2 = (B^\ast Bx)^\ast (B^\ast Bx) = x^\ast B^\ast BB^\ast Bx=x^\ast B^\ast B^\ast BBx=0.$
3. Therefore $B^\ast Bx=0$ and in turn $\|Bx\|^2 = x^\ast B^\ast Bx=0$ for all $x$.
4. So $Bx=0$ for all $x$. That is, $B=A^{k-1}=0$.

All the eigenvalues of a nilpotent matrix must be zero (this can be seen by taking powers of the Jordan canonical form). A normal matrix is diagonalizable. So $A=U \Lambda U^H$ where $\Lambda$ is the diagonal matrix containing the eigenvalues on the diagonal. But $\Lambda$ must be zero because $A$ is nilpotent. So $A=U 0 U^H=0$.

If you can use the spectral theorem then you know that $A$ is similar to a diagonal matrix $D$. Since $A$ is nilpotent, so is $D$. But then $D$ needs to be $0$.