In a complex inner product space a normal nilpotent linear map is the zero map.
Let the map be $T$. Since normal we have $\|T\| = r(T)$, the spectral radius. Again we have $r(T) = \lim\limits_{n \to \infty} \|T^n\|^{\frac1n} = 0$, since $T$ is nilpotent. Thus $\|T\| = 0$ and hence $T =0$.
Is the proof correct?
Another solution: You can prove this another way, too, by showing that, if $T$ is normal, then $\mathcal{N}(T)=\mathcal{N}(T^n)$ for all $n \ge 1$. To prove this, suppose $T$ is normal, and suppose that $T^2x=0$. Then \begin{align} &0=\langle T^2x,T^2x \rangle= \langle T^*Tx,T^*Tx\rangle \\ &\implies T^*Tx = 0 \\ &\implies 0 = \langle T^*Tx,x\rangle = \|Tx\|^2 \\ &\implies Tx=0. \end{align} Therefore $T^nx=0$ for some $n > 0$ iff $Tx=0$. In particular, the only nilpotent normal operator is the $0$ operator.