If $\kappa$ is weakly inaccessible, then it is a limit cardinal and hence $\kappa=\aleph_\lambda$ for some limit ordinal $\lambda$. Since the cofinality of $\aleph_\lambda$ is the same as the cofinality of $\lambda$, it follows by the regularity of $\kappa$ that $\lambda=\kappa$, and so $\kappa=\aleph_\kappa$, an $\aleph$-fixed point.
The next $\aleph$-fixed point after any ordinal $\beta_0$ must have cofinality $\omega$, since it is $\sup_n\beta_n$, where $\beta_{n+1}=\aleph_{\beta_n}$. So if a weakly inaccessible $\kappa$ is the $\delta$-th $\aleph$-fixed point, it cannot be that $\delta$ is a successor ordinal. But also, since and so $\delta$ is a limit ordinal. Since the $\aleph$-fixed points are closed, it also cannot be that this implies $\delta$ is a limit below \kappa$ has the same cofinality as $\kappa$, since then it would be singular, \delta$, and so by regularity it follows that $\kappa=\delta$ and thus, $\kappa$ must be is the $\kappa$-th fixed point.
Essentially the same argument works with $\beth$ and strongly inaccessible cardinals, simply by replacing $\aleph$ everywhere with $\beth$.
