Why are an even number of flips required to get back to the original list? 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".
Here is a combinatoric proof (which I am not sure if it is from permutation group).

Count the number of 'reversed pair' in the list. What I mean by reversed pair is pair of numbers which the one bigger in value comes before the smaller one. For example, in $[2,1,3,4]$, $(2,1)$ is a reversed pair but $(1,3)$ is not.

Now observe the change of the number of reversed pair when a flip is taken. Let's say the list is $[\cdots,a,\cdots,b,\cdots]$ and we flip $a,b$, turning it into $[\cdots,b,\cdots,a,\cdots]$. It should be obvious that those in front or behind does not contribute any changes. Also, for those in between, lets call one of it $c$, will only contribute if $a&lt;c&lt;b$ or $a&gt;c&gt;b$. $+2$ for the former and $-2$ for the latter. And finally, the $a,b$ pair. $+1$ if $a&lt;b$ and $-1$ if $a&gt;b$. Anyway, the total number changes by an odd number. Since you have total number of $0$ at beginning and you want $0$ at the end, you have to flip even number of times.

Hope this helps you.
The value of k will be positive since to get back two numbers back to its position we have to do two flips. 

For ex - for[1,2,3,4,5] we have first flip it once i.e.2 and 5 now to get back the original set again we again have to flip 2 and 5. 
Now if we use k flips then it must be even since the have to be reversed that are reversed by k-1 flips.