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Actually, the thing is to bypass $V=L$. So, is there a way to prove that (if consistent) $\mathsf{ZFC}$ can't prove that there exists a weakly inaccessible without first showing that $\mathsf{GCH}$ is relatively consistent? Obviously, if we can show that $\mathsf{ZFC}$ doesn't prove the consistency of a weakly inaccessible is much better.

The only really similar question I found is this, and I wasn't much too thorough in my web search because I'm pretty sure that this is just a curiosity.

PS. Answers using Easton magic are not allowed!

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    Does you accept using $L_\kappa$ directly? – 2017-01-15
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    @HanulJeon The point is to avoid the $L$ construction altogether. I don't know, perhaps some wise reflection trick... Good luck! – 2017-01-15

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