Let $Card$ be the proper class of all cardinals, define an infinite set of operators like $\otimes_{n}:(Card\setminus \omega)\times (Card\setminus\{0\})\longrightarrow Card$ which are defined for each natural number $n\geq 0$ recursively:
Definition 1: For $n=0$ define $\otimes_{0}$ as follows:
$\forall \kappa\geq \aleph_0~\forall \lambda>0~~~~~\kappa\otimes_{0}\lambda:=\kappa^\lambda$
If $\otimes_{n}$ is defined, consider $\otimes_{n+1}$ as follows:
Fix $\kappa\geq \aleph_0$, then define:
$\kappa\otimes_{n+1}1:=\kappa$
$\forall \lambda>0~~~~~\kappa\otimes_{n+1}\lambda^{+}:=(\kappa\otimes_{n+1}\lambda)\otimes_{n}\kappa$
$\forall \lambda>0~~~~$ if $\lambda$ is a limit cardinal then $\kappa\otimes_{n+1}\lambda:=\sup(\{\kappa\otimes_{n+1}\delta~|~\delta<\lambda\})$
Definition 2: An uncountable regular cardinal $\kappa$ is super inaccessible if for all $n\in \omega$ and for all $\lambda,\theta<\kappa$ we have $\lambda\otimes_{n}\theta<\kappa$ (i.e. $\kappa$ is closed under all operators $\otimes_{n}$)
Question 1: What is the consistency strength of the existence of a super inaccessible cardinal?
Question 2: Is every strongly inaccessible cardinal super inaccessible?
When I calculate these operations using your recursion equations, I get $\kappa\otimes_n\lambda=2^\kappa$ whenever $1\leq n<\omega$ and $\lambda\geq 2$. Am I misreading something, or is your definition not what you intended? Of course, if the definition is what you intended and I'm computing correctly, then "superinaccessible" is trivially equivalent to "inaccessible".