TITLE: Any hint about solving this monster determinant?
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: 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 :)