TITLE: Why are an even number of flips required to get back to the original list?
QUESTION [9 upvotes]: Consider the list of numbers $[1, \cdots, n]$ for some positive integer $n$. Two distinct elements $i$ and $j$ of the list can be switched in a so-called flip. For example, let $f$ be a flip that switches $2$ and $4$. Then $f([1,2,3,4]) = [1,4,3,2]$. Now consider a sequence of $k$ flips $f_1, \cdots, f_k$ of the list $[1, \cdots, n]$ such that $f_1(f_2(\cdots f_k([1,\cdots,n])\cdots)) = [1,\cdots, n]$, i.e. performing all flips gives the original list. Then $k$ must be even.
I would like to find a proof of this proposition that is elementary as possible. I already came up with a justification using permutation groups which goes as follows:
Each flip $f_i$ corresponds to a transposition of the list $[1, \cdots, n]$. Since the composition $f_1f_2\cdots f_k$ results in the identity, it must be an even permutation. Thus any representation of $f_1f_2\cdots f_k$  as a product of transpositions must contain an even number of transpositions. In particular, since each $f_i$ is a transposition, it follows that $k$ must be even. 
This "proof" trivializes the problem statement as it is by using relatively high-powered facts about permutation groups. Is there a lower-level proof that avoids using theses results? (Ideally such a proof would avoid permutation groups altogether and be understandable to the layman.) 
Edit: to clarify (since this question hasn't been getting as much attention as I had hoped), any proof that avoids re-deriving these powerful results about permutations would suffice. Basically I would want a proof that does not prove much more than the question requires, i.e. doesn't have a part where it says "in particular".

REPLY [3 votes]: Let $\ell=(x_1,x_2,\ldots, x_n)$ be a list of the numbers in $[n]$. For each $k\in[n]$ denote by $p_k$ the number of $x_i\ (1\leq i<k)$ with $x_i>x_k\,$, and call $\beta(\ell):=\sum_{k=1}^n p_k$  the badness of $\ell$. It is easy to convince oneself that a transposition ("flip") $\tau:\>\ell\mapsto\ell'$ changes the badness by $\pm1$. It follows that an odd number of flips cannot restore a given $\ell$ to its original version.