This is an exercise from Kunen:
Exercise I.13.17 If $\kappa$ is weakly inaccessible, then it is the $\kappa$th element of $\{\alpha: \alpha =\aleph_\alpha\}$. If $\kappa$ is strongly inaccessible, then it is the $\kappa$th element of $\{\alpha: \alpha =\beth_\alpha\}$.
I know that $\kappa$ is weakly inaccessible if and only if it is a regular $\aleph$-fixed point, and that $\kappa$ is strongly inaccessible if and only if it is a regular $\beth$-fixed point.
Moreover I know that we can enumerate the $\aleph$-fixed points and $\beth$-fixed points using normal functions $f_\aleph$ and $f_\beth$:
$$f_\aleph(0)= \text{smallest $\aleph$-fixed point}$$ $$f_\aleph(\alpha+1)= \text{smallest $\aleph$-fixed point greater than $\alpha$}$$ $$f_\aleph(\alpha)= \sup_{\beta<\alpha} f_\aleph(\beta) \quad \alpha \text{ nonzero limit}$$
and similarly for $f_\beth$. I also know that $f_\aleph(0)$ and $f_\beth(0)$ are not regular.
My strategy: Suppose $\kappa$ is weakly inaccessible. Then $\kappa=f_\aleph(\alpha)$ for some $\alpha$. It suffices to show that $\alpha$ is not a successor, for then $\alpha$ is nonzero limit, and so by regularity and the properties of normal functions:
$$\alpha\leq f_\aleph(\alpha)=\text{cf}(f_\aleph(\alpha))=\text{cf}(\alpha)\leq \alpha$$
i.e. $\kappa=f_\aleph(\alpha)=\alpha$.
How can I show that $\alpha$ is not a successor?
The trick here is to show that whatever $\mu$ may be, the least fixed point above it must have countable cofinality.
This is done by the usual proof that a fixed point exists. Start with $\lambda_0=\mu^+$ and $\lambda_{n+1} = \aleph_{\lambda_n}$. Then $\lambda = \sup\lambda_n$ is a fixed point, it has countable cofinality, and it is the least fixed point above $\mu$.