Forum:Question about Inaccessibles

I was wondering, is there another way to write \(\psi(\epsilon_{I+1})\) with the psi function? Does this equal \(\psi_{\Omega_{I+1}}(0)\) ? I know that \(\psi(\epsilon_{\Omega_x+1})\) = \(\psi(\psi_{x}(0))\) for all finite x, and I was wondering if this is true for all x, even when x is inaccessible.