Proving $V_{\kappa}$ is a model of ZFC for inaccessible $\kappa$ Prove that if $\kappa$ is an inaccessible cardinal, then $V_{\kappa}$ satisfies all the axioms of ZFC.
How is this done for the axiom of choice and for regularity?
 A: The idea is that you need to remember that whenever ZFC requires you something to be a set, you want it to be an element of $V_\kappa$, as this is the model.
So to expand Arturo's comment, the axiom of choice says that when you have a set of non-empty sets the product is non-empty. Since the sets of $V_\kappa$ are elements of it in the world, you can take a product of them, from the inaccessibility of $\kappa$ you'd have that the rank of the product (in the big universe) is still less than $\kappa$ and thus the product remains within $V_\kappa$ as you'd like it to.
As for regularity, well... basically the relation $\in$ is the same as in the world, so it's well-founded and everything holds.
A: Let's solve the general problem surrounding the question,
with a few observations, each of them easy to see.


*

*Every $V_\alpha$ for any ordinal $\alpha$ satisfies
Extensionality and Foundation, since all transitive sets satisfy
Extensionality and Foundation.

*Every $V_\alpha$ satisfies Separation, for the simple reason that
$A\subset B\in V_\alpha\implies A\in V_\alpha$.

*Every $V_\alpha$ satisfies Union, since $A\in
 V_\alpha\implies \bigcup A\in V_\alpha$.

*Every $V_\lambda$ for a limit ordinal $\lambda$
satisfies Pairing and Powerset, since the required set is
added at the next stage below $\lambda$.

*Every $V_\alpha$ satisfies the Axiom of Choice (assuming
this holds in $V$) in the choice-set version, since if
${\cal A}\in V_\alpha$ is a family of disjoint sets , then all choice
sets $B\subset \bigcup {\cal A}$ selecting one element from each $A\in{\cal A}$
have the same or lower rank than ${\cal A}$, and so $B\in
 V_\alpha$.

*Every $V_\alpha$ with $\omega\lt\alpha$ satisfies
Infinity, since $\omega\in V_\alpha$.

*The only remaining axiom is Replacement, and this is the
only one that makes use of inaccessibility. But if
$\kappa$ is inaccessible, then $V_\kappa$ satisfies
Replacement, since if $A\in V_\kappa$ and $F:A\to
 V_\kappa$ is definable over $V_\kappa$, then $F''A$ has
bounded rank below $\kappa$, since $|A|\lt\kappa$ and
$\kappa$ is regular. Thus, $F''A\in V_\kappa$ as desired.
Finally, one can also consider the question about
$H_\delta$, the sets of hereditary size less than $\delta$,
and things are a bit nicer here in several ways.


*

*For any regular uncountable cardinal $\delta$, the set $H_\delta$ of
sets having hereditary size less than $\delta$, satisfies
$ZFC^-$, that is, all of ZFC except the Powerset axiom.
One gets the easy axioms easily; Separation is easy since the subset also has small hereditary size;
and Replacment follows from the fact that the union of fewer
than $\delta$ many sets in $H_\delta$ still has size less
than $\delta$ by the regularity of $\delta$.


In particular, this shows that ZFC proves that there are
numerous transitive models of $ZFC^-$.


*

*If $\kappa$ is inaccessible, then $V_\kappa=H_\kappa$,
and this satisfies full ZFC, since we get $ZFC^-$ in
$H_\kappa$, and we get power set since $\kappa$ is a
strong limit.

