 Solution

For the intended infinite regular cardinal $\kappa$, set $S_0(\delta )=0$. Suppose $0<\zeta <{\kappa }^{+}$ and the functions for its predecessors have been constructed. Since $|\zeta |\le \kappa$, choose a surjection $e_{\zeta }:\kappa \to \zeta$ and defineThe supremum is below $\kappa$: it uses fewer than $\kappa$ ordinals below $\kappa$, and $\kappa$ is regular. This defines a function $\kappa \to \kappa$ at each stage of the recursion.

For any $\alpha <\zeta$, choose $\eta$ with $e_{\zeta }(\eta )=\alpha$. Whenever $\delta >\eta$, the displayed supremum includes $S_{\alpha }(\delta )+1$, so $S_{\alpha }(\delta )<S_{\zeta }(\delta )$. Therefore the recursion yields a long chain under eventual domination of length ${\kappa }^{+}$.