If ZFC has a model then the union of ZFC with the negation of inaccessible cardinals has a model

I would appreciate some help with proving the statement in the title.

I'm new to model theory, so I won't understand technical terms or symbols that well. Hence a sketch of the proof will suffice. Thanks!

Regarding the definition of 'inaccessible': A cardinal $\lambda$ is inaccessible iff

1. $\lambda$ is uncountable, i.e. $\lambda > \omega$,
2. $\lambda$ is a strong limit, i.e. for all $\mu < \lambda$, we have that $2^{\mu} < \lambda$ and
3. $\lambda$ is regular, i.e. for every $X \subseteq \lambda$ such that $\operatorname{card}(X) < \lambda$, we have that $\sup X < \lambda$.

Let $M$ be the model you start out with.
If $M$ has no inaccessible cardinals, then you're done.
Otherwise let $\kappa$ be the first inaccessible cardinal in $M$. Presumably you already know that the model's $V_\kappa$ is a model of ZFC, when viewed from $M$ itself and therefore also (why?) when viewed from outside the model.
You then just need to prove that $V_\kappa$ doesn't contain any inaccessible cardinal. (It is easy to get lost here and say that this is immediate because $\kappa$ was the first inaccessible. But you need to prove that there is no member of $V_\kappa$ that $V_\kappa$ itself thinks is inaccessible, even if $M$ doesn't).
• Looking back at this exercise I'm a little confused about the last step, i.e. proving that $V_{\kappa}$ doesn't contain any inaccessible cardinal. I'm also a little confused about the meaning of "when viewed from $M$ itself and therefore also when viewed from outside the model.". I'm sorry about this late remark, but I went on vacation and accepted your answer without going through the details properly. – Laplace's Demon Sep 4 '16 at 14:36