How to evaluate the determinant of this $n \times n$ matrix?

Question

Find the following determinant: $$\begin{vmatrix} 0 &1&0&0 &\cdots& 0\\ 0 &0&1&0 &\cdots& 0\\ 0 &0&0&1 &\cdots& 0 \\ \vdots & \vdots &\cdots&\cdots&\cdots&\vdots\\0&0&0&0 &\cdots&1\\1&0&0&0 &\cdots&0\end{vmatrix}_{n \times n}$$

Attempt:
I switched $1^{st}$ row with the last one, then second with the last, and so on till the $(n-1)^{th}$ row with the $n^{th}$ (last) row and there I had all $1$'s in the diagonal so I came up with the answer $1$.

Am I correct? Can anyone solve this exercise please?

Answer 1

Hint: row exchanges change the value of the determinant by a factor of $-1$, and the determinant of the identity is $1$. How do you make your matrix the identity by exchanging rows?

Answer 2

The corresponding matrix is the companion matrix for the polynomial $$f(x)=x^n-1$$ So the characteristic polynomial for this matrix is $f(x)$. So determinant is same as product of all $n$ th root of unity, which is $(-1)^{n+1}$

Answer 3

Expand along the first column. $\Delta=(-1)^{n+1}|I_{n-1}|=(-1)^{n+1}$, where $I_{n-1}$ is the identity matrix of order $n-1$.