simplified formula of expression having adjoint of a matrix Well, the question is :
let A be a square matrix of order $n_n$ and adj(A) is the adjoint of matrix A
then is it possible to get a simplified formula for the expression
$$adj(adj(adj(...adj(A))))$$ here  adj() has been written 'r' times.
The reason I am asking this question is that there exists a simplified formula for the determinant of the given expression if $|A|$ represents determinant of a matrix A then
$$|adj(adj(adj(...adj(A))))| = |A|^{(n-1)^r}$$
This can be easily proved using the identity
$$A*adj(A) = |A|*I_n$$
for $r=2$ I know that
$$adj(adj(A)) = |A|^{n-2}*A$$ but I can't generalize this result. So, is it possible to get a general form of this expression? All help is greatly appreciated.
 A: Let $A$ be $n\times n$.
When $n=1$, $\operatorname{adj}(A)$ is equal to $1$ by convention. In turn, the $m$-fold adjugate matrix of $A$ is always equal to $1$. Note that the identity $\operatorname{adj}\left(\operatorname{adj}(A)\right)=\det(A)^{n-2}A$ only applies to the case where $n\ge2$.
When $n=2$, $\operatorname{adj}\left(\operatorname{adj}(A)\right)=A$. Hence the $m$-fold adjugate of $A$ is $\operatorname{adj}(A)$ when $m$ is odd, or $A$ when $m$ is even.
Now suppose $n>2$. If $A$ is singular, $\operatorname{adj}\left(\operatorname{adj}(A)\right)=0$. Therefore the $m$-fold adjugate of $A$ is $\operatorname{adj}(A)$ when $m=1$, or zero when $m>1$.
If $n>2$ and $A$ is nonsingular, we have, for every integer $k$,
\begin{aligned}
\operatorname{adj}(\det(A)^kA)
&=\operatorname{det}\left(\det(A)^kA\right)\left(\det(A)^kA\right)^{-1}
=\det(A)^{(n-1)k+1}A^{-1},\\
\operatorname{adj}(\det(A)^kA^{-1})
&=\operatorname{det}\left(\det(A)^kA^{-1}\right)\left(\det(A)^kA^{-1}\right)^{-1}
=\det(A)^{(n-1)k-1}A.
\end{aligned}
It follows that the $m$-fold adjugate matrix of $A$ is
$$
\operatorname{adj}^m(A)=\begin{cases}
\det(A)^{a_m}A^{-1}&\text{when $m\ge1$ is odd},\\
\det(A)^{a_m}A&\text{when $m\ge1$ is even}.\\
\end{cases}
$$
where $a_m$ is given by the recurrence relation
\begin{aligned}
a_1&=1,\\
a_2&=n-2,\\
a_m
&=\begin{cases}
(n-1)a_{m-1}+1&\text{when $m\ge3$ is odd},\\
(n-1)a_{m-1}-1&\text{when $m\ge3$ is even},\\
\end{cases}\\
&=\begin{cases}
(n-1)\left[(n-1)a_{m-2}-1\right]+1&\text{when $m\ge3$ is odd},\\
(n-1)\left[(n-1)a_{m-2}+1\right]-1&\text{when $m\ge3$ is even},\\
\end{cases}\\
&=\begin{cases}
(n-1)^2a_{m-2}-(n-2)&\text{when $m\ge3$ is odd},\\
(n-1)^2a_{m-2}+(n-2)&\text{when $m\ge3$ is even},\\
\end{cases}\\
&=\begin{cases}
(n-1)^{m-1}-\frac{(n-1)^{m-1}-1}{(n-1)^2-1}(n-2)&\text{when $m$ is odd},\\
\frac{(n-1)^m-1}{(n-1)^2-1}(n-2)&\text{when $m$ is even}.\\
\end{cases}\\
\end{aligned}
