# Why $\mathrm{rank}(x^y) < \alpha+\omega$, if $x$, $y$ have rank $\le$ $\alpha$?

This question is from Set Theory, Jech(2006), Page 70, 6.5.

Rank function is defined as on Page 64:

• $V_0=\emptyset$,
• $V_{\alpha+1}=P(V_{\alpha})$,
• $V_{\alpha}=\bigcup_{\beta<\alpha}V_\beta$, if $\alpha$ is a limit ordinal.

$\mathrm{rank}(x)=\operatorname{min}\{\alpha \in \mathrm{Ord}:x \in V_{\alpha+1}\}$

If $x , y \in V_{\alpha + 1}$, then for each $u \in x$ and $v \in y$ we have that $u , v \in V_\alpha$ and so $\{ v \} , \{ v , u \} \in V_{\alpha + 1}$ and so $\langle v , u \rangle = \{ \{ v \} , \{ v , u \} \} \in V_{\alpha + 2}$. Thus $y \times x \subseteq V_{\alpha + 2}$, and so $y \times x \in V_{\alpha + 3}$. Also, every subset of $y \times x$ belongs to $V_{\alpha + 3}$, in particular every function $y \to x$ belongs to $V_{\alpha + 3}$, and so the set of all those functions belongs to $V_{\alpha + 4}$.
Show that the rank of $y\times x$ is below $\alpha+5$ or so, and the conclusion should follow.