MHB Proving Even # of Transpositions for Identical Permutations

[simo1]

is there any easier way of proving that no matter how an identical permutation say (e) is written the number of transpositins is even.

my work
i tried let t_1...t_n be m transpositions then try to prove that e can be rewritten as a product of m-2transpositions.
i had x be any numeral appearing in one of the transpositions t_1...t_n where t_k=(xa) and t_k is the last transposition in e=t_1t_2...t_m. i tried this and it seems very long:(

 

[Deveno]

That is actually a difficult thing to prove, although it "seems" obvious it should be true.

This is the "standard" proof:

Consider the expression:

$\displaystyle m = \prod_{1\leq i < j \leq n} (i - j)$.

We define:

$\displaystyle \sigma(m) = \prod_{1 \leq i < j < \leq n} (\sigma(i) - \sigma(j))$ for $\sigma \in S_n$.

Note that if $\sigma$ is a transposition, that $\sigma(m) = -m$. Also note that no matter what permutation $\sigma$ is, we have:

$\sigma(m) = \pm m$.

Note also that if $\sigma = \tau\pi$ we have:

$\displaystyle\sigma(m) = \tau(\pi(m)) = \prod_{1 \leq i < j < \leq n} (\tau(\pi(i)) - \tau(\pi(j)))$

Now if a permutation $\sigma$ could be written as both an even number and odd number of transpositions, we obtain:

$m = -m \implies m = 0$, which is impossible, since all the factors of $m$ are non-zero (An odd number of transpositions changes the sign of $m$ an odd number of times, resulting in $-m$, and an even number of transpositions changes the sign of $m$ an even number of times, resulting in no change to $m$).