about the smallest $k$ such that $V_k$ is a model of ZFC Let $k$ be the smallest ordinal such that $V_k$ is a model of ZFC. I know that $k$ need not be inaccessible cardinal, and $k$ has cofinality $\omega$.
Then how big is $k$? How can we write down $k$ in terms of $\aleph$ or $\beth$? Since its confinality is $\omega$, how can we find an $\omega$-sequence to reach $k$?
 A: These cardinals are now called worldly. As Asaf pointed out, any such $\kappa$ is a beth fixed point. Note that $V_\kappa$ cannot see any $\omega$-sequence cofinal in $\kappa$, since $\mathsf{ZFC}$ proves that no $\omega$-sequence of ordinals is cofinal in the class $\mathsf{ORD}$ of all ordinals. This means that any witnessing sequence cannot be "easily" described, or else $V_\kappa$ would be able to identify it.
The following argument may help you understand why the cofinality of the smallest such cardinal is $\omega$: Let $\mathsf{ZFC}_n$ be the subtheory of $\mathsf{ZFC}$ resulting from restricting the axiom schema of replacement to $\Sigma_n$ formulas. Let $C_n$ be the class of cardinals $\kappa$ such that $V_\kappa$ models $\mathsf{ZFC}_n$. By the reflection theorem, each $C_n$ is cofinal in $\mathsf{ORD}$. In fact, since $\Sigma_n$ satisfaction is definable, each $C_n$ contains a club: Consider those $\kappa$ such that $V_\kappa$ is a $\Sigma_n$-elementary substructure of the universe $V$.
Suppose $\kappa$ is worldly. [Note that, since satisfaction is not definable -- this is Tarski's theorem, -- although we know, from outside, that each $C_n^{V_\kappa}$ contains a club in $\kappa$, in $V_\kappa$ we do not have access to the sequence $(C_n^{V_\kappa}\mid n<\omega)$.] If $\kappa$ has cofinality larger than $\omega$, then $\bigcap_n C_n^{V_\kappa}$ again contains a club in $\kappa$. If $\rho$ is any element of this intersection, then $\rho$ is again worldly, since it satisfies $\Sigma_n$ replacement for all $n$. In fact, if we take $\rho$ in the intersection of the clubs contained in the $C_n^{V_\kappa}$, then we have the stronger conclusion that $V_\rho\prec V_\kappa$.
It follows that no such $\kappa$ can be the first worldly cardinal, and therefore the smallest one has cofinality $\omega$ (and this is witnessed by the fact that there is an $\omega$-sequence of clubs with empty intersection). Easy modifications of the argument show that actually a very long initial segment of the sequence of worldly cardinals consists solely of cardinals of cofinality $\omega$. The first worldly cardinal of uncountable cofinality has cofinality $\omega_1$, the next one again has cofinality $\omega$, and again all worldly cardinals past this one have cofinality $\omega$ for a long stretch, then we see again one of cofinality $\omega_1$, etc.
A: Indeed the least such $\kappa$, if it exists, has a countable cofinality. However $\kappa$ is a $\beth$-fixed point. This means that $\kappa=\beth_\kappa$. So in particular it is a bit hard to write a cofinal sequence in an explicit form.
And note that every $\beth$-fixed point is an $\aleph$-fixed point: $\beth_\alpha\geq\aleph_\alpha\geq\alpha$ is provable in $\sf ZFC$. If $\alpha=\beth_\alpha$ then we have equality all across the board.
All we can do is prove that this cardinal has a countable cofinality, and therefore such sequence exists.
