Can you tell me if my answer is correct:
Show that the set $P^{\omega_\omega}(\omega)$ exists.
My answer:
Let $P^0 (\omega) = \omega$, $P^{\alpha + 1}(\omega) = P(P^\alpha (\omega))$ and for a limit ordinal $\lambda$ let $P^\lambda (\omega) =\bigcup_{\beta < \lambda} P^\beta (\omega)$.
The transfinite recursion theorem tells us that if $G: V \to V$ is a class function from the class of all sets to the class of all sets then there exists a unique function $F: ON \to V$ with $F(\alpha) = G(F\mid_\alpha)$.
Hence to show the existence of $P^{\omega_\omega}(\omega)$ we need to define a $G: V \to V$ with $G(F\mid_{\omega_\omega}) = P^{\omega_\omega}(\omega) = F(\omega_\omega)$.
Also we know that for successor ordinals $\alpha$ we want $F\mid_{\beta + 1} = F(\beta) = P^{\beta}(\omega)$. So for successor ordinals $\alpha = \beta + 1$ we define $G(F\mid_{\alpha})= G(F\mid_{\beta + 1}) = G( P^{\beta}(\omega)) := P( P^{\beta}(\omega))$.
We also know that $F\mid_\varnothing = \varnothing$ so $G(F\mid_\varnothing) = G(\varnothing) := P^0 (\omega)$.
Finally, if $\lambda$ is a limit ordinal we want $G(F\mid_\lambda) := \bigcup_{\alpha < \lambda} P^\alpha (\omega)$.
For all other sets we define $G$ to map to the empty set.
Thanks for your help.