Determinant of matrix whose $(i,j)$ entry is $\min(i,j)$ Let $n \in\mathbb{N}$ and define $A \in M_{n}(\mathbb{R})$ by $A(i,j)= \min(i,j)$ for $i,j \in \{1, 2 ,3, 4,\cdots, n\}$.Compute $\det(A)$.
My try is with a example:
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.
Help, please.

By looking at your example, it seems that by using Laplace expansion in the last column, you would get twice your induction assumption. Let me ellaborate: your determinant would be
$$-1\left|\begin{array}{ccc} 1 & 2 & 2 \\
1 & 2 & 3 \\
1 & 2 & 3\end{array}\right|+2\left|\begin{array}{ccc} 1 & 1 & 1 \\
1 & 2 & 3 \\
1 & 2 & 3\end{array}\right|-3\left|\begin{array}{ccc} 1 & 1 & 1 \\
1 & 2 & 2 \\
1 & 2 & 3\end{array}\right|+4\left|\begin{array}{ccc} 1 & 1 & 1 \\
1 & 2 & 2 \\
1 & 2 & 3\end{array}\right|$$
The first two determinants are $0$, since there are repeated rows, and the last two are equal to $1$ by your induction assumption, so it is $4-3=1$. I am pretty sure that this approach Works in the general $n\times n$ matix, so try it out and see what happens.
