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".