Does the following define a Mahlo cardinal? Let M be a cardinal with the following properties:
  - M is regular
- $\kappa &lt; M \implies 2^\kappa &lt; M$
- $\kappa &lt; M \implies s(\kappa) &lt; M$ where $s(\kappa)$ is the smallest strongly inaccessible cardinal strictly greater than $\kappa$  

My question is: Is M a Mahlo cardinal ? If so, how does the definition above connect to the usual definition in terms of stationary sets ?

Motivation: My intuition about a Mahlo cardinal is that it you cannot reach it by taking unions, power sets or "the next inaccessible cardinal", which is my definition above.
My worry is, I might have arrived at something much smaller than Mahlo.
No, your condition doesn't imply Mahloness.

First, note that your first two conditions simply state that $M$ is inaccessible, and the third one gives that $M$ is limit of inaccessibles. Now consider the first cardinal $M_0$ which is inaccessible limit of inaccessibles. That is, every inaccessible below $M_0$ is not a limit of inaccessibles. Now consider $C$ consisting of limits of the set $I$ of inaccessible cardinals smaller than $M_0$. By choice of $M_0$, $C$ contains no inaccessibles below $M_0$, so if we show $C$ is a club below $M_0$, we will get that $M_0$ is not Mahlo.

$C$ is closed: this is true because the set of limit points of any set is closed. For the sake of completion, I will sketch the argument: suppose that each of $\alpha_i$ is a limit point of $I$, where $(\alpha_i)_{i&lt;\kappa}$ is an increasing sequence of cardinals with limit $A$. Let $\beta&lt;A$. This means that there is some $\alpha_i$ between $\beta$ and $A$. Since $\beta&lt;\alpha_i$, there is some element of $I$ which is between $\beta$ and $\alpha_i$, hence between $\beta$ and $A$. So $A$ is a limit point of $I$, so $A\in C$. Hence $C$ is closed.

$C$ is unbounded below $M_0$: Let $\alpha&lt;M_0$. Then $s(\alpha),s(s(\alpha)),s(s(s(\alpha))),...$ is an increasing sequence in $I$ of length $\omega$. Since cofinality of $M_0$ is (clearly) greater than $\omega$, limit of this sequence, which is an element of $C$, is between $\alpha$ and $M_0$. Hence $C$ is unbounded below $M_0$.

Indeed, the condition "inaccessible limit of inaccessibles" (or simply "regular limit of inaccessibles") is a condition known as 1-inaccessibility. It can be generalized: we can speak of 2-inaccessibles, which are regular limits of 1-inaccessibles, 3-inaccessibles which are regular limits of 2-inaccessibles and so on. This can be obviously extended to a successor ordinal, and for limit ordinal $\alpha$ (e.g. $\alpha=\omega$) we say that cardinal is $\alpha$-inaccessible if it's $\beta$-inacessible for $\beta&lt;\alpha$ (this isn't circular definition, but rather it proceeds by transfinite induction).

For every ordinal $\alpha$, notion of $\alpha+1$-inaccessibility is strictly stronger than that of $\alpha$-inaccessibility. As can be shown, every Mahlo cardinal $\kappa$ is not only 1-inaccessible, 2-inaccessible, $\omega$-inaccessible, but even $\kappa$-inaccessible. Notion of Mahloness is even stronger than that, but for that, I'd recommend reading this section on Wikipedia article.
Thanks a lot for your answers, Wojowu and Andreas.
I find I am not able to simply reply with a comment while being a guest, so I am writing it this way.

The mental picture I am getting of Mahlo versus inaccessibles is somewhat like $\aleph_1$ versus the Veblen hierarchy of countable ordinals - you keep recursively strengthening definitions and taking fixed points, but just never get there. Don't know how strong the analogy is, though.