If A is a square matrix of order n then $\underset{(n-1)times}{\underbrace{\left(adjadjadj\mathrm{....}adjA\right)}}.\underset{ntimes}{\underbrace{\left(adjadjadj\mathrm{....}adjA\right)}}$ is equal to

We know $A\left(adjA\right)=\left|A\right|I_n$

$\left(adjA\right)\left(adj\left(adjA\right)\right)=\left|adjA\right|I_n={\left|A\right|}^{n-1}.I_n$

$adj\left(adjA\right)\left(adjadjadjA\right)=\left|adjadj\left(A\right)\right|I_n={\left|A\right|}^{{(n-1)}^2}{I}_n$

$\Rightarrow \underset{n-1times}{\underbrace{\left(adjadjadj\mathrm{...}adjA\right)}}\underset{ntimes}{\underbrace{\left(adjadjadj\mathrm{...}adjA\right)}}={\left|A\right|}^{{(n-1)}^{(n-1)}}{I}_n$