Determinant of matrix with $A_{ij} = \min (i, j)$ Given a $n\times n$ matrix whose $(i, j)$-th entry is the lower of $i,j$, eg.
$$\begin{pmatrix}1 & 1 & 1 & 1\\
1 & 2 & 2 & 2 \\
1 & 2 & 3 & 3\\
1 & 2 & 3 & 4 \end{pmatrix}.$$
The determinant of any such matrix is $1$. 
How do I prove this?
Tried induction but the assumption would only help me to compute the term for $A_{nn}^*$ mirror.
 A: You can substract the $j$-th column to the $(j+1)$-th one. This will leave you with a lower-triangular matrix of all ones.
A: Rotate the matrix A by $180$ degree. Do you think the determinant would change? Have a look at this question if you have any doubt about this.
Does rotating a matrix change its determinant?
Let R denote the rotated matrix
Let's have a look at the inverse of this matrix $R$
$$D_{n} = (R_{n})^{-1}=
        \begin{pmatrix}
        1 & P^{T} \\
        P &  B_{n-1}
        \end{pmatrix}
$$ where, $$B =
        \begin{pmatrix}
        2 & -1 & 0 & \cdots &0 \\
        -1 & 2 & -1 & \cdots & 0 \\
        0&-1&2&\cdots &0\\
        \vdots & \vdots &\vdots &\ddots&\vdots\\   
        0 & 0 & 0&\cdots& 2 \\
        \end{pmatrix}
$$
and $P = (-1,0,0,0...,0)^{T}$
You may find it a good exercise to verify this.
By expanding the determinant along row $1$ we have $\forall n\geq 2$
$$det(D_{n}) = det(B_{n-1})-det(B_{n-2})$$ where,
B follows the recursion formulla $$det(B_{n}) = 2det(B_{n-1})-det(B_{n-2})$$ Base case $det(B_{1}) = 2$, which gives $det(B_{n}) = n+1. $
Hence, For such matrices D we have $det(D_{n}) = 1 \implies det(R_{n}) = det(A_{n}) = 1 $
