T is a normal operator; prove that R(T) = R(T*)

Question: Let $T$ is a normal operator on a finite-dimensional inner product space $V$. Show that $N(T)=N(T^*)$ and that $R(T)=R(T^*)$.
I think that I got the first part. Suppose $x \in N(T)$ Then $T(x)=0$. But $||T(x)|| = ||T^*(x)||$ for all $x$ so $T^*(x)=0$ as well. So $x \in N(T^*)$, which means that $N(T) \subseteq N(T^*)$. The same logic goes the other way around and we conclude $N(T) = N(T^*)$.
But what about $R(T)=R(T^*)$? I know that $R(T^*) = N(T)^{perp}$ but how does that help?
$R(T^*) = N(T)^{perp}=N(T^*)^{perp}=R(T)$