After reading about various properties of $V_\alpha$ and how it can be used to model various axioms of Set Theory, Kunen mentions that in $ZFC$, one cannot prove that there is an $\alpha$ such that $V_\alpha \models ZFC$ (assuming $ZFC$ is consistent.) I've seen proofs before that show why certain axioms fail under certain $V_\alpha$'s, but I was intrigued by the strength of this statement.
The Incompleteness theorem is then mentioned to show why this is true by contratiction. Assuming $ZFC \vdash \exists \alpha \thinspace[V_\alpha \models ZFC]$. Then, $ZFC \vdash \mbox{Con}(ZFC)$, contradicting the Second Incompleteness Theorem.
I was suprised to see the instance of a "nicer" contradiction. Assuming $ZFC \vdash \exists \alpha \thinspace[V_\alpha \models ZFC]$, let $\beta$ be least such that $V_\beta \models ZFC$. But then, by our assuption, $V_\beta \models \exists \alpha[V_\alpha \models ZFC]$. Such an $\alpha \in V_\beta$ must be less than $\beta$, which contradicts $\beta$ being the least.
However, to finish the "nicer" contradiction, it must be verified that the statement "$V_\alpha \models ZFC$" is absolute. How would one go about showing that this statement is absolute? Any help/hints would be greatly appreciated.
Assuming $\mathsf{ZFC} \vdash \exists \alpha\,(V_\alpha \models \mathsf{ZFC})$, let $\beta$ be the least ordinal such that $V_\beta \models \mathsf{ZFC}$ and let $\alpha$ be the unique set such that the model $V_\beta$ satsifies "$\alpha$ is the least ordinal such that $V_\alpha$ satisfies $\mathsf{ZFC}$." Let $X = V_\alpha^{V_\beta}$.
It suffices to show that the following statements, true in $V_\beta$, are absolute between $V_\beta$ and $V$.
$\alpha$ is an ordinal,
$X \models \mathsf{ZFC}$, and
$X = V_\alpha$.
For 1 and 2 we only need to use the fact that $V_\beta$ is a transitive set and that $\Delta_0$ formulas are absolute to transitive sets. For 3, although the property of being a rank initial segment of $V$ is not absolute to all transitive sets, one can show that it is absolute to rank initial segments of $V$. One way to do this is to show by induction on $\xi$ that $(V_\xi)^{V_\beta} = V_\xi$ for all ordinals $\xi < \beta$.