Prove that if $A$ is a normal operator on an complex separable hilbert space $H$ and $A^{3} = A^{4}$ then $A$ is self adjoint

Prove that if $A$ is a normal operator on an complex separable hilbert space $H$ and $A^{3} = A^{4}$ then $A$ is self adjoint

I am not quite sure how to do this proof. My first idea was to use the spectral theorem for normal operators to construct a proof similar to to finite dimensional case.

This is also true if we drop the separability condition.

If you know the basics about $C^*$-algebras, there is an easy proof by a standard technique.

Consider $B:= C^*(A,1)$, the unital $C^*$-algebra generated by the operator $A$ in $B(H)$. Since $A$ is a normal operator, $B$ is a commutative $C^*$-algebra, hence $B \cong C(X)$ for a compact Hausdorff space $X$. But in $C(X)$, the statement is trivial: indeed if a function $f\in C(X)$ satisfies $f^3 = f^4$, then $f$ can only take the values $0$ or $1$ and in particular $f$ is real-valued, so self-adjoint. Through the isomorphism $B\cong C(X)$, you find that $A$ must also be self-adjoint, as desired.