Any hint about solving this monster determinant? I'm asked to solve the following determinant:
$$|A|=
\begin{vmatrix}
1 &2 &3 &\cdots &{n-1} &n\\
2 &3 &4 &\cdots &n &1\\
\vdots &\vdots &\vdots & &\vdots &\vdots\\
{n-1} &n &1 &\cdots &{n-3} &{n-2}\\
n &1 &2 &\cdots &{n-2} &{n-1}
\end{vmatrix}
$$
My attempt is to add all the other columns onto the first one, which gives
$$|A|=\frac{n(n+1)}{2}|B|$$
where $|B|$ is, however, none the easier than $|A|$.
I think the result should be very special, since $A$ is a very special symmetric matrix itself. But I simply get stuck. Can you help me? thanks in advance.

EDIT 
It just occurred to me that definition might work out well here. Am I on the right track?
I'm now into another question. If $(j_1,j_2,\cdots,j_n)$ is an $n-th$ permutation of ${1,2,\cdots,n}$ and the number of inversion pairs in there is $\tau$, then what's the number of inversion pairs in its inverse permutation $(j_n,j_{n-1},\cdots,j_2,j_1)$ ? This may shed a light on the problem.
Some friend of mine has given me a relatively simple solution, which I will add subsequently  as an answer. 
 A: $A$ is symmetric and therefore has $n$ orthogonal real eigenvectors with eigenvalue $\lambda_1,\ldots, \lambda_n$ and we have $\det A=\lambda_1\cdot\ldots\cdot \lambda_n$. Guessing eigenvectors might help. Unfortunately this guessing  (apart from the "all ones" vector)  is only really simple in the complexification:
If $\zeta\in\mathbb C$ is an $n$th root of unity, let $v_\zeta:=(1,\zeta,\zeta^2,\ldots,\zeta^{n-1})^T$.
Observe from the fact that multiplication by $\zeta$ rotates $v_\zeta$ componentwise that $$Av_\zeta=(1+2\zeta+\ldots+n\zeta^{n-1})\cdot v_{\zeta^{-1}}.$$
Hence $v_\zeta$ is an eigenvector of $A^2$ with eigenvalue  $|1+2\zeta+\ldots+n\zeta^{n-1}|^2$. Since the $v_\zeta$ are linearly independant, we conclude that 
$$(\det A)^2= \prod_{\zeta^n=1}|1+2\zeta+\ldots+n\zeta^{n-1}|^2$$
hence
$$\tag1\det A =\pm \prod_{\zeta^n=1} |1+2\zeta+\ldots+n\zeta^{n-1}|$$
Let's simplify: 
We have
$$ \tag2(1+2\zeta+\ldots+n\zeta^{n-1})(1-\zeta)=1+\zeta+\ldots+\zeta^{n-1}-n$$
and 
$$ \tag3(1+\zeta+\ldots+\zeta^{n-1})(1-\zeta)=1-\zeta^n=0.$$
If $\zeta=1$ then clearly $1+2\zeta+\ldots+n\zeta^{n-1}=\frac{n(n-1)}{2}$. In all other cases, $(2)$ and $(3)$ imply $$(1+2\zeta+\ldots+n\zeta^{n-1})=\frac n{\zeta-1}.$$
Since the $\zeta-1$ with $\zeta^n=1$ and $\zeta\ne 1$ are precisely the roots of the polynomial $$\frac{(X+1)^n-1}{X}=X^{n-1}+\ldots + n$$
we conclude that $n=(-1)^{n-1}\prod_{\zeta^n=1,\zeta\ne 1}(\zeta-1)$.
Therefore $(1)$ becomes
$$|\det A|=\left|\frac{n(n+1)}{2}\cdot \frac{n^{n-1}}{\prod_{\zeta^n=1, \zeta\ne 1}(\zeta-1)}\right|=\frac{n^{n-1}(n+1)}{2}.$$
I am still fighting with the sign of $\det A$, though ...
A: Another way of looking at the $\det(A)$:
by swapping pairs of rows $(1,n-1),(2,n-2)\dots$
this matrix is transformed into 
circulant matrix $C$ with the same $0$-th row $(1,\dots,n)$,
which has a well known explicit formula for determinant,
\begin{align}
\det(C)&=\prod_{j=0}^{n-1}\sum_{k=0}^{n-1} (k+1)\exp\left(2\pi i\frac{jk}{n}\right).
\end{align}
Accounting to the number of swapped rows,
\begin{align}
\det(A)&=(-1)^{\lfloor (n+3)/2\rfloor}\det(C).
\end{align}
And as @Michael Biro has pointed out in the comment,
A052182 gives 
\begin{align}
\det(C)&=(-1)^{n-1}\frac{(n+1)n^{n-1}}{2},
\end{align}
hence
\begin{align}
\det(A)&=(-1)^{\lfloor n/2\rfloor}\frac{(n+1)n^{n-1}}{2}.
\end{align}
A: A friend of mine has given me a relatively easy solution as follows. ( Apologies that I'm on the mobile device so it's not convenient to code)

Well, I just uploaded the picture from my PC. It looks okay now :)
