The reason why the existence of inaccessible cardinals cannot be proven in ZFC I am curious about why the existence of inaccessible cardinals cannot be proven in ZFC.My intuitive proof is:
Suppose we can prove there is an inaccessible cardinal ϰ,then by definition we must show that for any cardinals b&c which satisfying b<ϰ and c<ϰ,then b^c<ϰ.Then if we consider the set S which is the union of all the cardinals less than ϰ and consider the power set of S--P(S).It follows that |P(S)|>=ϰ.This implies we can proof the existence of inaccessible cardinals under ZFC iff we can disprove it.Therefore we cannot prove the existence using ZFC.
Is this the intuitive idea behind the proof?
I only have learnt a little bit in set theory by self-studying.But I don't know any thing about mathematical logic.So I don't know the thing such as the second incompleteness theorem.This statement was found in my set theory book but without prrof it.So any comment and improvement of my work is welcome!

The simplest proof for why we cannot prove the existence of inaccessible cardinals is as follows:
If $\kappa$ is inaccessible, then $V_\kappa$ is a model of $\sf ZFC$. Now if the existence of inaccessible cardinals was provable, look inside $V_\kappa$, where $\kappa$ is the least inaccessible cardinal. There you should find some $\lambda$ which is inaccessible as well. However this would imply $\lambda$ is also inaccessible, which is a contradiction since $\kappa$ was the minimal. 
