Define: $\operatorname {wo}^n(x) \iff \forall y (y \in^n x \to \operatorname {wo} (y))$

Where $\operatorname {wo}(y)$ refers to $y$ being well orderable.

Where $y \in^0 x \iff y=x \\ y \in^{n+1} x \iff \exists z (z \in^n x \land y \in z)$

**n-well ordered choice:** for $n=0,1,2,...$, for every set $x$ of nonempty sets, if $\operatorname{wo}^n(x)$, then $x$ admits a choice function on it.

If we add this schema to axioms of $\sf ZF$, would it entail axiom of choice?