If $A$ is normal and upper triangular then it is diagonal 
Let $A$ be a normal matrix in Mat$_{n\times n}(\mathbb C)$, if $A$ is upper triangular then it is diagonal

(Normal means $AA^*=A^*A$, where $A^*$ is the conjugate transpose of $A$)
If I consider the diagonal of $AA^*$,  let denote $(a_{ij})=A$ and $(â_{ij})_{i,j}=AA^*$ then, since $AA^*=A^*A$
$â_{ii}=\sum\limits_{k=1}^na_{ik}\overline{a}_{ik}=\sum\limits_{k=1}^n\overline{a_{ki}}{a}_{ki}$
$\implies\sum\limits_{k=1}^n|a_{ik}|^2=\sum\limits_{k=1}^n|a_{ki}|^2$. 
If I take $i=n$ then it follows that $a_{in}=0, \forall 1\le i\le n-1$ and continuing in this manner the upper diagonal entries are zero, Is this correct ?
Can I show it in another way, because in a previous exercise I had to show that ''If A is normal and nilpotent then $A=0$'' so using this can I decompose $A$ into diagonal and nilpotent matrix, then show that the nilpotent part is zero ?

Denote $A=\begin{pmatrix}a_{11}&a_{12}&\cdots&a_{1n}\\
&a_{22}&\cdots&a_{2n}\\&&\ddots&\vdots\\&&&a_{nn}\end{pmatrix}$. 
Observe that the $(1,1)$-entries of $A^\ast A$ and $AA^\ast$ are
$$\vert a_{11}\vert^2\quad\mbox{and}\quad \sum_{i=1}^n\vert a_{1i}\vert^2,$$
respectively. Since $A$ is normal,
$$
\sum_{i=1}^n\vert a_{1i}\vert^2=\vert a_{11}\vert^2\quad\Rightarrow\quad 
\sum_{i=2}^n\vert a_{1i}\vert^2=0\quad\Rightarrow\quad a_{12}=a_{13}=\cdots=a_{1n}=0.$$
Now, from preceding result, the $(2,2)$-entry of $A^\ast A$ is
$$\vert a_{12}\vert^2+\vert a_{22}\vert^2=\vert a_{22}\vert^2,$$
and the $(2,2)$-entry of $AA^\ast$ is 
$\displaystyle\sum_{i=2}^n\vert a_{2i}\vert^2$. Again, since $A$ is normal,
$$
\sum_{i=2}^n\vert a_{2i}\vert^2=\vert a_{22}\vert^2\quad\Rightarrow\quad 
\sum_{i=3}^n\vert a_{2i}\vert^2=0
\quad\Rightarrow\quad a_{23}=a_{24}=\cdots=a_{2n}=0.$$
Continue this process, we may conclude that the upper off-diagonal entries of $A$ are all zero. Hence, $A$ is a diagonal matrix.
