minimum-maximum entries matrix Let $M(n)$ be an $n\times n$ matrix in the variables $x_1,\dots,x_n$ with entries
$$M_{i,j}(n)=\frac{x_{\max(i,j)}}{x_{\min(i,j)}}, \qquad 1\leq i,j\leq n.$$
I'm interested in the following:

Questions.
(1) Is there a neat or "closed form" evaluation for the determinant $\det M(n)$?
(2) Is there an explicit formula for the inverse of $M(n)$?

Thank you.
 A: I wish to advertise the Method of Condensation in proving determinantal evaluation. The key is guess the answer, which is thanks to Ehud Meir. Then, generalize it a bit. Let $x_1, x_2,\dots$ be an infinite set of variables and modify the original matrix (by shifting variables) to $M^{a,b}(n)$ so that
$$M_{i,j}^{a,b}(n):=\frac{x_{\max(i+a,j+b)}}{x_{\min(i+a,j+b)}}.$$
Convention: $M^{0,0}(n)=M(n)$.  
Claim. If $a\neq b$ then $\det M^{a,b}(n)=0$, and if $a=b$ then
$$\det M^{a,a}(n)=\prod_{r=2}^n\frac{x_{k-1+a}^2-x_{k+a}^2}{x_{k-1+a}^2}.\tag1$$
Proof. The case $a\neq b$ is easy - simply factor out a variable from $n^{th}$-column/row and another variable from the $(n-1)^{th}$-column/row. These new columns/rows are identical.
Inductive proofs neatly work with this Dodgson's recursive relation
$$\det Z^{0,0}(n)=\frac{\det Z^{1,1}(n-1)\det Z^{1,1}(n-1)-\det Z^{0,1}(n-1)\det Z^{1,0}(n-1)}{\det Z^{1,1}(n-2)}$$
satisfied by any matrix (so long as denominators do not vanish). Thus, it holds for $\det M^{a,b}(n)$.
So, it remains to prove that the (explicit) formula on the RHS of (1) does satisfy the same equation. However, this is quite a routine simplification (preferably with symbolic sofwares). The proof follows. $\square$
A: Let us write $$a_r=\frac{x_{r+1}}{x_r}$$ for $r=1\cdots n$. 
We can then write the matrix $M(n)$ in the form 
$$\begin{pmatrix} 1 & a_1 & a_1a_2& \cdots & a_1a_2\cdots a_{n-1} \\ a_1 & 1 & a_2& \cdots & a_2\cdots a_{n-1}\\ \vdots & \vdots & \vdots &\ddots & \vdots\\ a_1a_2\cdots a_{n-1}& a_2\cdots a_{n-1} &\cdots & a_{n-1} & 1\end{pmatrix} $$ 
We do now Gauss elimination, and reduce $a_{n-1}$ times the $(n-1)$-th row from the $n$-th row. We then get in the last row $0, 0, \ldots (1-a_{n-1}^2)$.
But this means that $det(M(n))=det(M(n-1))(1-a_{n-1}^2)$ and by induction
$$det(M(n)) = \prod_{r=1}^{n-1} (1-a_r^2)=\prod_{r=1}^{n-1}(1-\frac{x_{r+1}^2}{x_r^2}).$$
By using inductively the Gauss elimination mentioned above, one can also get the inverse of $M(n)$.
