Is this permutation even or odd? Here is the question that I am working on:

Let $\sigma$ be the permutation of the numbers $1,2,...,n$ which reverses their order completely. That is,
$$\sigma=\begin{pmatrix} 1 & 2 & 3 &...&n \\ n & n-1 & n-2&...&1  \end{pmatrix}$$
  Is $\sigma$ even or odd? 

Here is what I noticed. In general, if I want to find whether a permutation is even or odd, I can write down the permutation in disjoint cycle form and then express that as a composition of transpositions. So, for example, $(123)$ would be even because $(123)$ = $(13)(12)$. The problem is that I'm not sure if this approach can apply to my original question since the permutation $\sigma$ depends on the number $n$. Any more insight on this question would be helpful.
EDIT As explained by the users below, I initially misinterpreted the question, so disregard my first comments in the chat below.
 A: It is hard to write the permutations neatly so I will use words instead.  
Note that the first and last elements, $1$ and $n$, are just interchanged.  Similarly, the second and second to last, $2$ and $n - 1$, are interchanged, $3$ and $n - 2$, etc.  
If $n$ is even then every element is swapped and there are $\frac{n}{2}$ 2-cycles.  So, if $\frac{n}{2}$ is even then the permutation is even and if $\frac{n}{2}$ is odd then the permutation is odd.  
If $n$ is odd then the element in the middle, $\frac{n+1}{2}$ will be fixed.  The remaining $n-1$ elements will be swapped by $\frac{n-1}{2}$ 2-cycles.  
So summarising:
$n = 0 \mod 4$ No fixed element and the permutation is even.
$n = 1 \mod 4$ Middle element is fixed and the permutation is even.
$n = 2 \mod 4$ No fixed element and the permutation is odd.
$n = 3 \mod 4$ Middle element is fixed and the permutation is odd.  
A: If $n$ is even, $\sigma=(1\; n)(2\; n-1) \cdots \left(\frac{n}{2} \; \frac{n}{2}+1\right)$.
If $n$ is odd, $\sigma = (1\; n)(2\; n-1) \cdots \left(\frac{n-1}{2}\; \frac{n+1}{2}\right)$.
Then count the number of transpositions. (Yes, it will depend on $n$.)
