Return to Answer

Just one clarification to Guillaume's answer: Yes, a model of ZFC is a set $E$ together with a binary relation $R$ on $E$ such that $(E,R)$ satisfies ZFC. The relation $R$ is extensional, that is every $x\in E$ is determined by the collection of $y\in E$ with $yRx$, because $(E,R)$ satisfies the axiom of extensionality.
However, even though $(E,R)$ satisfies the axiom of regularity, the relation $R$ need not be well-founded. $(E,R)$ thinks that $R$ is well-founded, but the real world knows an infinite $R$-decreasing chain in $E$. In this situation, $(E,R)$ is a non-standard model of ZFC which is clearly not isomorphic to a transitive model with the $\in$-relation.

However, an inaccessible cardinal $\kappa$ actually gives you a transitive model of ZFC, and this is more than just having any model of ZFC. (And most approaches to forcing would like to have a transitive set that is a model of ZFC (with the binary relation being the usual $\in$-relation). But there are reasons why we can pretend that we have a transitive set model of ZFC even if we don't really have one.) Now, it is not the case that the inaccessibility of $\kappa$ somehow miraculously gives you a new set that is a model of ZFC. It is more the case that $\kappa$ is so large that if you cut off the universe at $\kappa$, this initial part of the full set-theoretic universe already satisfies ZFC.
I.e., from the point of view of $V_\kappa$, $\kappa$ already looks like the class of all ordinals.

What makes this work is the fact that an inaccessible cardinal has excellent closure properties. $\forall\lambda<\kappa(2^\lambda<\kappa)$ guarantees that every element of $V_\kappa$ is of size $<\kappa$ and the regularity of $\kappa$ now gives you the axiom of replacement. The other axioms are rather easily satisfied in this context.

Note, however, that an inaccesible cardinal can cease to be inaccessible (or even a cardinal) after enlarging the universe (for example by forcing).

Just one clarification to Guillaume's answer (his answer has been edited by now): Yes, a model of ZFC is a set $E$ together with a binary relation $R$ on $E$ such that $(E,R)$ satisfies ZFC. The relation $R$ is extensional, that is every $x\in E$ is determined by the collection of $y\in E$ with $yRx$, because $(E,R)$ satisfies the axiom of extensionality.
However, even though $(E,R)$ satisfies the axiom of regularity, the relation $R$ need not be well-founded. $(E,R)$ thinks that $R$ is well-founded, but the real world knows an infinite $R$-decreasing chain in $E$. In this situation, $(E,R)$ is a non-standard model of ZFC which is clearly not isomorphic to a transitive model with the $\in$-relation.

However, an inaccessible cardinal $\kappa$ actually gives you a transitive model of ZFC, and this is more than just having any model of ZFC. (And most approaches to forcing would like to have a transitive set that is a model of ZFC (with the binary relation being the usual $\in$-relation). But there are reasons why we can pretend that we have a transitive set model of ZFC even if we don't really have one.) Now, it is not the case that the inaccessibility of $\kappa$ somehow miraculously gives you a new set that is a model of ZFC. It is more the case that $\kappa$ is so large that if you cut off the universe at $\kappa$, this initial part of the full set-theoretic universe already satisfies ZFC.
I.e., from the point of view of $V_\kappa$, $\kappa$ already looks like the class of all ordinals.

What makes this work is the fact that an inaccessible cardinal has excellent closure properties. $\forall\lambda<\kappa(2^\lambda<\kappa)$ guarantees that every element of $V_\kappa$ is of size $<\kappa$ and the regularity of $\kappa$ now gives you the axiom of replacement. The other axioms are rather easily satisfied in this context.

Note, however, that an inaccesible cardinal can cease to be inaccessible (or even a cardinal) after enlarging the universe (for example by forcing). The $V_\kappa$ of the ground model (the small original universe) is still a model of ZFC, but the $V_\kappa$ of the enlarge universe (where $\kappa$ is not inaccessible anymore) need not be a model of ZFC.

Just one clarification to Guillaume's answer: Yes, a model of ZFC is a set $E$ together with a binary relation $R$ on $E$ such that $(E,R)$ satisfies ZFC. The relation $R$ is extensional, that is every $x\in E$ is determined by the collection of $y\in E$ with $yRx$, because $(E,R)$ satisfies the axiom of extensionality.
However, even though $(E,R)$ satisfies the axiom of regularity, the relation $R$ need not be well-founded. $(E,R)$ thinks that $R$ is well-founded, but the real world knows an infinite $R$-decreasing chain in $E$. In this situation, $(E,R)$ is a non-standard model of ZFC which is clearly not isomorphic to a transitive model with the $\in$-relation.

However, an inaccessible cardinal $\kappa$ actually gives you a transitive model of ZFC, and this is more than just having any model of ZFc.
  Now, it is not the case that the inaccessibility of $\kappa$ somehow miraculously gives you a new set that is a model of ZFC. It is more the case that $\kappa$ is so large that if you cut off the universe at $\kappa$, this initial part of the real universe already satisfies ZFC.
I.e., from the point of view of $V_\kappa$, $\kappa$ already looks like the class of all ordinals.

Just one clarification to Guillaume's answer: Yes, a model of ZFC is a set $E$ together with a binary relation $R$ on $E$ such that $(E,R)$ satisfies ZFC. The relation $R$ is extensional, that is every $x\in E$ is determined by the collection of $y\in E$ with $yRx$, because $(E,R)$ satisfies the axiom of extensionality.
However, even though $(E,R)$ satisfies the axiom of regularity, the relation $R$ need not be well-founded. $(E,R)$ thinks that $R$ is well-founded, but the real world knows an infinite $R$-decreasing chain in $E$. In this situation, $(E,R)$ is a non-standard model of ZFC which is clearly not isomorphic to a transitive model with the $\in$-relation.

However, an inaccessible cardinal $\kappa$ actually gives you a transitive model of ZFC, and this is more than just having any model of ZFC. (And most approaches to forcing would like to have a transitive set that is a model of ZFC (with the binary relation being the usual $\in$-relation). But there are reasons why we can pretend that we have a transitive set model of ZFC even if we don't really have one.) Now, it is not the case that the inaccessibility of $\kappa$ somehow miraculously gives you a new set that is a model of ZFC. It is more the case that $\kappa$ is so large that if you cut off the universe at $\kappa$, this initial part of the full set-theoretic universe already satisfies ZFC.
I.e., from the point of view of $V_\kappa$, $\kappa$ already looks like the class of all ordinals.

What makes this work is the fact that an inaccessible cardinal has excellent closure properties. $\forall\lambda<\kappa(2^\lambda<\kappa)$ guarantees that every element of $V_\kappa$ is of size $<\kappa$ and the regularity of $\kappa$ now gives you the axiom of replacement. The other axioms are rather easily satisfied in this context.

Note, however, that an inaccesible cardinal can cease to be inaccessible (or even a cardinal) after enlarging the universe (for example by forcing).

Just one clarification to Guillaume's answer: Yes, a model of ZFC is a set $E$ together with a binary relation $R$ on $E$ such that $(E,R)$ satisfies ZFC. The relation $R$ is extensional, that is every $x\in E$ is determined by the collection of $y\in E$ with $yRx$, because $(E,R)$ satisfies the axiom of extensionality.
However, even though $(E,R)$ satisfies the axiom of regularity, the relation $R$ need not be well-founded. $(E,R)$ thinks that $R$ is well-founded, but the real world knows an infinite $R$-decreasing chain in $E$. In this situation, $(E,R)$ is a non-standard model of ZFC which is clearly not isomorphic to a transitive model with the $\in$-relation.

Just one clarification to Guillaume's answer: Yes, a model of ZFC is a set $E$ together with a binary relation $R$ on $E$ such that $(E,R)$ satisfies ZFC. The relation $R$ is extensional, that is every $x\in E$ is determined by the collection of $y\in E$ with $yRx$, because $(E,R)$ satisfies the axiom of extensionality.
However, even though $(E,R)$ satisfies the axiom of regularity, the relation $R$ need not be well-founded. $(E,R)$ thinks that $R$ is well-founded, but the real world knows an infinite $R$-decreasing chain in $E$. In this situation, $(E,R)$ is a non-standard model of ZFC which is clearly not isomorphic to a transitive model with the $\in$-relation.

However, an inaccessible cardinal $\kappa$ actually gives you a transitive model of ZFC, and this is more than just having any model of ZFc.
Now, it is not the case that the inaccessibility of $\kappa$ somehow miraculously gives you a new set that is a model of ZFC. It is more the case that $\kappa$ is so large that if you cut off the universe at $\kappa$, this initial part of the real universe already satisfies ZFC.
I.e., from the point of view of $V_\kappa$, $\kappa$ already looks like the class of all ordinals.