How do you find the determinant of the following matrix? \begin{pmatrix} 1&2&3&\cdots&n\\2&3&\cdots&n&1\\3&\cdots&n&1&2\\\vdots&\vdots&\vdots&\vdots&\vdots\\n&1&\cdots&(n-2)&(n-1) \end{pmatrix}
(i.e. we have the first row and all the other rows are cyclic shifts of the first one)
I've tried subtracting columns from one another to get a nice triangle matrix, but after some time that seemed futile for me. What I've noticed is that whichever anti-diagonal from right to left you choose, e.g., a diagonal
\begin{pmatrix} \require{cancel} \cdots&\cdots&\cancel{3}&\cdots\\\cdots&\cancel{3}&\dots&\dots\\\cancel{3}&\dots&\dots&\dots \end{pmatrix}
all of the numbers on this diagonal are similar (e.g., $(2,2), (3,3,3), (n, \dots, n)$ and so on), so maybe one can get a recurrence relation.