Circular determinant problem

Since the matrix with determinant $\Delta$ is circulant, its eigenvalues $\{\lambda_j\}_{j=1}^n$ are given by

$$\lambda_j=1+2\omega_j+3\omega_j^2+\ldots+n\omega_j^{n-1},$$

where $\omega_j=\exp\left(\frac{2\pi ij}{n}\right)$. The determinant is the product of the eigenvalues.

For an explicit formula not involving products and sums, see sequence A052182 in OEIS. It can be shown that

$$\Delta(n)=(-1)^{n-1} \cdot n^{n-2} \cdot \frac{n^2+n}{2}$$

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.