ZFC can't prove the existence of an ordinal $\alpha$ such that $V_\alpha\vDash {\rm ZFC}$ Given an ordinal $\alpha$ we will denote by $V_\alpha$ the $\alpha$-stage of the Von Neumann hierarchy for the set-theoretic universe. An exercise from Kunen's book says that ZFC can't prove the existence of an ordinal $\alpha$ such that $V_\alpha$ is a model of ZFC. This is clear by virtue of Gödel's Incompleteness Theorem but Kunen proposes a different approach, pointing out in a hint that if the existence of such an ordinal is provable, then taking $\alpha$ to be least such that ${\rm ZFC}\vdash "V_\alpha\vDash {\rm ZFC}"$, there would exist another ordinal $\beta\in \alpha$ such that ${\rm ZFC}\vdash "V_\beta\vDash {\rm ZFC}"$. 
Could someone explain to me why would there be such a $\beta$? Hints or any comments would be much appreciated. Thanks in advance.
 A: Suppose that ZFC proves the existence of such an $\alpha$, and take the least such $\alpha$.  Then $V_\alpha$ is a model of ZFC, so $V_\alpha\vDash\text{"There exists an ordinal $\beta$ such that $V_\beta$ is a model of ZFC"}$.  So there is some $\beta\in V_\alpha$ such that $$V_\alpha\vDash\text{"$\beta$ is an ordinal and $V_\beta$ is a model of ZFC"}.$$  If we knew that that last statement in quotes was absolute for $V_\alpha$, then we would conclude that $\beta$ is an ordinal and $V_\beta$ is a model of ZFC.  And since $\beta\in V_\alpha$, $\beta<\alpha$, so this contradicts the minimality of $\alpha$.  So what you need to prove is that "$\beta$ is an ordinal and $V_\beta$ is a model of ZFC" is absolute for $V_\alpha$.
(By the way, your phrasing "taking $\alpha$ the less ordinal such that $ZFC\vdash V_\alpha\vDash ZFC$" doesn't make much sense.  We don't want the least $\alpha$ such that ZFC proves that $V_\alpha$ is a model of ZFC (that doesn't make sense, because ZFC can't even express the sentence "$V_\alpha$ is a model of ZFC" for any particular ordinal $\alpha$ unless $\alpha$ is definable).  What we want is just the least $\alpha$ such that $V_\alpha\vDash ZFC$, without reference to what ZFC can prove.)
