Circular determinant problem I'm stuck in this question:
How calculate this determinant ?
$$\Delta=\left|\begin{array}{cccccc}
1&2&3&\cdots&\cdots&n\\
n&1&2&\cdots&\cdots& n-1\\
n-1&n&1&\cdots&\cdots&n-2\\
\vdots &\ddots & \ddots&\ddots&&\vdots\\
\vdots &\ddots & \ddots&\ddots&\ddots&\vdots\\
2&3&4&\cdots&\cdots&1
\end{array}\right|$$
Thanks a lot.

I'll do the case $n=4$ and leave the general case for you.
It is useful to recall how elementary row operations effect the value of determinant, see for example
ProofWiki.
We get
$$
\begin{vmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 4 & 1 \\
3 & 4 & 1 & 2 \\
4 & 1 & 2 & 3
\end{vmatrix}
\overset{(1)}=
\begin{vmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 4 & 1 \\
3 & 4 & 1 & 2 \\
10 & 10 & 10 & 10
\end{vmatrix}=
10\begin{vmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 4 & 1 \\
3 & 4 & 1 & 2 \\
1 & 1 & 1 & 1
\end{vmatrix}
\overset{(2)}=
10\begin{vmatrix}
0 & 1 & 2 & 3 \\
0 & 1 & 2 &-1 \\
0 & 1 &-2 &-1 \\
1 & 1 & 1 & 1
\end{vmatrix}
\overset{(3)}=
10\begin{vmatrix}
0 & 0 & 0 & 4 \\
0 & 0 & 4 & 0 \\
0 & 1 &-2 &-1 \\
1 & 1 & 1 & 1
\end{vmatrix}
=10\cdot 4^2
$$
(1): added first three rows to the last one
(2): subtracted a multiple of the 4th row from the other rows
(3): subtracted 2nd row from the 1st one, 3rd row from the 2nd one
There are several possibilities how to see that the determinant of the last matrix is $4^2$. For example, you can use Laplace expansion several times, until you get $2\times2$ matrix; you can use the expression of the determinant using permutations (the only permutation which will give a non-zero summand gives $a_{14}a_{23}a_{32}a_{41}$ for this matrix); or you can exchange rows in such way that you get upper triangular matrix.
