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
- $\lambda$ is uncountable, i.e. $\lambda > \omega$,
- $\lambda$ is a strong limit, i.e. for all $\mu < \lambda$, we have that $2^{\mu} < \lambda$ and
- $\lambda$ 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).