$\DeclareMathOperator{\adj}{adj}$
If $|\adj(A)| = |A|^{n - 1} ;\bigl| \adj\bigl(\adj(A)\bigr) \bigr| = {\left| A \right|^{{{\left( {n - 1} \right)}^2}}}$, then $\bigl| {\underbrace {\adj\dots\adj\bigl(\adj(A) \bigr)}_{t \text{ - times}}} \bigr| = {\left| A \right|^{{{\left( {n - 1} \right)}^t}}}$
How do we prove it?
Upto $\bigl|\adj\bigl(\adj(A)\bigr)\bigr| = | A |^{(n - 1)^2}$, it is mentioned in the book but not after that.
$\DeclareMathOperator{\adj}{adj}$ $$A(\adj A)=|A|I$$ Put $A=\adj\adj A$ $$(\adj\adj A)(\adj \adj\adj A)=|\adj\adj A|I$$ Now, $\adj\adj A=|A|^{n-2}A$. And, $|\adj\adj A|=|A|^{(n-1)^2}$. Putting this, we get, $$|A|^{n-2}A(\adj \adj\adj A)=|A|^{(n-1)^2}I$$ $$\implies \adj\adj\adj A=|A|^{n^2+1-2n-n+2}A^{-1}$$ $$=|A|^{n^2-3n+3}\frac{\adj A}{|A|}$$ $$=|A|^{n^2-3n+2}A$$ Using the fact that $|kA|=k^n|A|$, we get $|\adj\adj\adj A|=|A|^{(n-1)^3}$. You see the pattern now?