Find the determinant of order $n$:
$D_n=\begin{vmatrix} 0&1&2\ldots&n-1\\ n-1&0&1\ldots&n-2\\ n-2&n-1&0\ldots&n-3\\ \vdots &\vdots &\ddots &\vdots\\ 1&2&3\ldots&0 \end{vmatrix}$
I've tried developing it using Laplace's method, adding rows/columns, but it wasn't very helpful. Please give me some hints, thank you!
Here's an elementary solution to the problem.
Consider matrixs $J_{k}$ defined by $$J_{k}=\begin{bmatrix}0 & I_{n-k}\\I_{k}&0\end{bmatrix}$$
All the $J_{k}$ span a $\mathbb{C}$-linear space of which $\mathcal{B}=\{J_{k}:k = 0,\cdots,n-1\}$ is a basis. Denote the space as $V$, then $D_{n}\in V$. Furthermore, we get $J_{k} = J_{1}^{k}$ by calculation, thus for every $A\in V$
$$A = a_{0}I_{n} + a_{1}J_{1}+a_{2}J_{1}^{2}+\cdots+a_{k-1}J^{k-1}_{1} = g(J_{1})$$
Where $g$ is a polynomial of $k-1$ degree.
For $k=1$, the eigenpolynomial of $J_{1}$ is $\vert\lambda I_{n}-J_{1}\vert = \lambda^{n}-1$, indicating that $J_{1}$ has eigenvalues of $\{1,\omega,\omega^{2},\cdots,\omega^{n-1}\}$. Denote $\omega^{k}$ as $\omega_{k}$, then the eigenvector $v_{k}$ related to $\omega_{k}$ can also be calculated $$v_{k} = \begin{bmatrix}1\\\omega_{k}\\\omega_{k}^{2}\\\vdots\\\omega_{k}^{n-1} \end{bmatrix}$$ Take $P = [v_{0},v_{1},\cdots,v_{n-1}]^{T}$, we have $$PJ_{1}P^{-1}=\mathrm{\mathop{diag}}\{\omega_{0},\cdots,\omega_{n-1}\}$$ So for any $A = g(J_{1})$, we have $$PAP^{-1} = Pg(J_{1})P^{-1} = g(PJ_{1}P^{-1}) = \mathrm{\mathop{diag}}\{g(\omega_{0}),g(\omega_{1}),\cdots,g(\omega_{n-1})\}$$ Thus $\det A = \prod_{i=0}^{n-1}g(\omega_{i})$.
Take $A$ to be $D_{n}$, then you get the answer.
