Let $\Omega$ be a bounded simply connected domain in $\mathbb{C}$ with a boundary which is a Jordan curve with smoothness $C^{n,\alpha}$, where $n\geq1$, $0<\alpha<1$. The Kellogg Warschawski theorem (Thm 3.6 in Pommerenke - Boundary behaviour of Conformal maps) states that if $\Phi: \mathbb{D} \to \Omega$ is a conformal surjective map, then the $n^{th}$ derivative $\Phi^{(n)}$ extends continuously to the boundary with the extension being in the holder class $C^{\alpha}$. I am interested in the converse. More precisely:
Question: Fix $n\geq 1$, $0<\alpha<1$. Let $\Omega$ be a domain in $\mathbb{C}$ bounded by a Jordan curve. It is given that $\Phi^{(n)}$ extends continuously to the boundary with the extension being in the holder class $C^{\alpha}$. What can we say about the smoothness of the boundary of $\Omega$?
Pommerenke in his book states that if $\Phi^{(n)}$ extends continuously to the boundary for all $n$, then $\Omega$ is smooth. I definitely believe this statement but his proof is that we can just take the conformal parametrization of the curve which is smooth and hence the curve is smooth. This is fallacious as we do not know if the derivative of the parametrization is nonzero. For e.g. consider a curve in $\mathbb{C}$ which is parametrized conformally and locally at the origin looks like $z^{1.5}$. Then its derivative behaves like $z^{0.5}$, which becomes zero at $z=0$. My guess is that $\Omega$ should have smoothness of class $C^{n-1,\alpha}$ though I am not sure. Any help is appreciated.