In my linear algebra class, my professor gave us this determinant for practice:
$$det\pmatrix {1 & 2 & 3 & \dots & n \\ 2 & 3 & 4 & \dots & 1 \\ \vdots & \vdots & \vdots & \dots & \vdots \\ n & 1 & 2 & \dots & n-1}$$
Where the $i$-th row or column is the set $$\{1,2,\dots,n\}$$ shifted $i-1$ times to the left.
He recommended we add rows $2$ through $n$ to row $1$. And then factor out the constant. This yields:
$$\frac{n(n+1)}{2}det\pmatrix {1 & 1 & 1 & \dots & 1 \\ 2 & 3 & 4 & \dots & 1 \\ \vdots & \vdots & \vdots & \dots & \vdots \\ n & 1 & 2 & \dots & n-1}$$
However, I cannot for the life of me find a useful way to simplify this further. I have tried adding the first column to each subsequent column, subtracting the first row from each subsequent row, but nothing seems to simplify the matrix. Perhaps, there is an inductive solution, but I haven't been able to find one simply from the $2\times2$ and $3\times3$ cases.
Any help would be appreciated.
We use the fact that multiplying row $i$ by $\lambda \neq 0$ then adding it to row $j$ does not change the determinant of the matrix.
Also, multiplying row $i$ by $\lambda \neq 0$ also multiples the determinant by $\lambda$. Thus,
\begin{align} \ & det\pmatrix{1 & 2 & 3 & \dots & n \\ 2 & 3 & 4 & \dots & 1 \\ 3 & 4 & 5 & \dots & 2 \\ \vdots & \vdots & \vdots & & \vdots \\ n & 1 & 2 & \dots & n-1} \\ \ & = \frac{n(n+1)}{2} det\pmatrix{1 & 1 & 1 & \dots & 1 \\ 2 & 3 & 4 & \dots & 1 \\ 3 & 4 & 5 & \dots & 2 \\ \vdots & \vdots & \vdots & & \vdots \\ n & 1 & 2 & \dots & n-1} && \text{as you have} \\ \ & = \frac{n(n+1)}{2} det\pmatrix{1 & 1 & 1 & \dots & 1 & 1 & 1 \\ 0 & 1 & 2 & \dots & n-3 & n-2 & -1 \\ 0 & 1 & 2 & \dots & n-3 & -2 & -1 \\ \vdots & \vdots & \vdots & & \vdots & \vdots & \vdots \\ 0 & 1 & 2-n & \dots & -3 & -2 & -1\\0 & 1-n & 2-n & \dots & -3 & -2 & -1} \\ \ & = \frac{n(n+1)}{2} det\pmatrix{1 & 1 & 1 & \dots & 1 & 1 & 1 \\ 0 & 1 & 2 & \dots & n-3 & n-2 & -1 \\ 0 & 0 & 0 & \dots & 0 & -n & 0 \\ 0 & 0 & 0 & \dots & -n & -n & 0 \\ \vdots & \vdots & \vdots & & \vdots & \vdots & \vdots \\ 0 & 0 & -n & \dots & -n & -n & 0\\0 & -n & -n & \dots & -n & -n & 0} \\ \ & = \frac{n(n+1)}{2} det\pmatrix{1 & 1 & 1 & \dots & 1 & 1 & 1 \\ 0 & 1 & 2 & \dots & n-3 & n-2 & -1 \\ 0 & 0 & 0 & \dots & 0 & -n & 0 \\ 0 & 0 & 0 & \dots & -n & 0 & 0 \\ \vdots & \vdots & \vdots & & \vdots & \vdots & \vdots \\ 0 & 0 & -n & \dots & 0 & 0 & 0\\0 & -n & 0 & \dots & 0 & 0 & 0} \\ \ & = \frac{(-1)^nn^{n-1}(n+1)}{2} det\pmatrix{1 & 1 & 1 & \dots & 1 & 1 & 1 \\ 0 & 1 & 2 & \dots & n-3 & n-2 & -1 \\ 0 & 0 & 0 & \dots & 0 & 1 & 0 \\ 0 & 0 & 0 & \dots & 1 & 0 & 0 \\ \vdots & \vdots & \vdots & & \vdots & \vdots & \vdots \\ 0 & 0 & 1 & \dots & 0 & 0 & 0\\0 & 1 & 0 & \dots & 0 & 0 & 0} \\ \ & = \frac{(-1)^nn^{n-1}(n+1)}{2}\cdot (1 \cdot 1 \cdot 1 \cdot \; \cdots \cdot 1 \cdot -1)(-1)^{\lceil n/2 \rceil +1} \\ \ & = \frac{(-1)^{n+\lceil n/2 \rceil}n^{n-1}(n+1)}{2} \end{align}
where the second to last line uses the permutation formula for the determinant.
The permutation formula gives the stated result because, as I have explained in the comment section, you have to choose one entry from each row and column and compute their product.
But in the last row, we have to choose the $1$ in order to not get a product of $0$. Similarly with the second to last row, up to the third row.
So far, we have already chosen one entry from each of columns $2$ to $n-1$, thus we need to choose an entry from columns $1$ and $n$ in the remaining two rows. In the second row, we can only choose the $-1$ at the far right to avoid a $0$ product, meaning we can only choose the $1$ in the first column for the first row.
This product is $1\cdot 1\cdot 1\dots 1\cdot -1 \cdot 1=-1$
The parity of this permutation is even if $n=1,2,5,6,9,10,\dots$ and is odd if $n=3,4,7,8,11,12,\dots$, so $sgn (\sigma) = (-1)^{\lceil n/2 \rceil +1}$, giving the determinant of the reduced matrix to be $-1 \cdot (-1)^{\lceil n/2 \rceil +1} = (-1)^{\lceil n/2 \rceil}$