Determinant formed by cofactors We are given delta(0)

And let delta(1) denote the determinant formed by the cofactors of elements of delta(0) and delta(2) denote the determinant formed by  cofactors of delta(1). Then we have to find value of delta(n) in term of delta(0). 
I tried it a lot . But does not got any start . 
 A: For this we first need to know the concept of Reciprocal Determinant. It is defined as: When in a given determinant, each element is replaced by it's cofactor, then the determinant so formed is called reciprocal determinant of the given determinant. If the original determinant is $D$, then the reciprocal determinant is given by $D'$.

Theorem: If $D$ is a determinant of order $n$, and $D'$ be its reciprocal determinant then $D' =D^{n-1}$.  

Proof: Let
    $$ D=
  \det{\begin{vmatrix}
    a_1 & b_1 & \cdots & n_1\\
    a_2 & b_2& \cdots & n_2\\
    \vdots & \vdots & \ddots & \vdots \\
    a_n & b_n & \cdots & n_n\\\
\end{vmatrix}}
$$ and 
    $$ D'=
  \det{\begin{vmatrix}
    A_1 & B_1 & \cdots & N_1\\
    A_2 & B_2& \cdots & N_2\\
    \vdots & \vdots & \ddots & \vdots \\
    A_n & B_n & \cdots & N_n\\\
\end{vmatrix}}
$$ where capital letters denote the cofactors of the small letters. Also, we know that $a_iA_j+b_iB_j+...n_iN_j=D$ where $i=j$ and $a_iA_j+b_iB_j+...n_iN_j=0$ when $i\neq j$. Thus on multiplying $D$ and $D'$, we have, 
    $$ DD'=
    \det{\begin{vmatrix}
    D & 0 & \cdots & 0\\
    0 & D & \cdots & 0\\
    \vdots & \vdots & \ddots & \vdots \\
    0 & 0 & \cdots & D\\\
\end{vmatrix}} =D^{n}
$$. Thus $D'=D^{n-1}$. Hope you can then take it from here on successive repetitions of this theorem.


A: |D1|=|Cd0| =|D0|^n-1                           { properties=|adj.[a]|=|a|^n-1} here order is 3
|D2|=|Cd1|=|D0|^(n-1)^2
|D3|=|Cd2|=|D0|^(n-1)^3
|Dn|=|Cd(n-1)=|D0|^(3-1)^n
|Dn|=|Cd(n-1)=|D0|^2n
|Dn|=|D0|^2n
