Example: Showing That Mahlo Cardinals Are Hyper-inaccessible
Suppose κ is Mahlo. We proceed by transfinite induction on α to show that κ is α-inaccessible for any α ≤ κ. Since κ is Mahlo, κ is inaccessible; and thus 0-inaccessible, which is the same thing.
If κ is α-inaccessible, then there are β-inaccessibles (for β < α) arbitrarily close to κ. Consider the set of simultaneous limits of such β-inaccessibles larger than some threshold but less than κ. It is unbounded in κ (imagine rotating through β-inaccessibles for β < α ω-times choosing a larger cardinal each time, then take the limit which is less than κ by regularity (this is what fails if α ≥ κ)). It is closed, so it is club in κ. So, by κ's Mahlo-ness, it contains an inaccessible. That inaccessible is actually an α-inaccessible. So κ is α+1-inaccessible.
If λ ≤ κ is a limit ordinal and κ is α-inaccessible for all α < λ, then every β < λ is also less than α for some α < λ. So this case is trivial. In particular, κ is κ-inaccessible and thus hyper-inaccessible.
To show that κ is a limit of hyper-inaccessibles and thus 1-hyper-inaccessible, we need to show that the diagonal set of cardinals μ < κ which are α-inaccessible for every α < μ is club in κ. Choose a 0-inaccessible above the threshold, call it α0. Then pick an α0-inaccessible, call it α1. Keep repeating this and taking limits at limits until you reach a fixed point, call it μ. Then μ has the required property (being a simultaneous limit of α-inaccessibles for all α < μ) and is less than κ by regularity. Limits of such cardinals also have the property, so the set of them is club in κ. By Mahlo-ness of κ, there is an inaccessible in this set and it is hyper-inaccessible. So κ is 1-hyper-inaccessible. We can intersect this same club set with the stationary set less than κ to get a stationary set of hyper-inaccessibles less than κ.
The rest of the proof that κ is α-hyper-inaccessible mimics the proof that it is α-inaccessible. So κ is hyper-hyper-inaccessible, etc..
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