What about the identity? (It is the product of any non-negative even number of transpositions.)

@Shaun Well $(1) = (a_1 a_2) (a_1 a_2)$ since a transposition is its own inverse. So does that mean this proposition does not apply for $n$ less than $3$?

Just pick an even number greater than $n-1$.

The identity is the empty product of transpositions, so it's a product of 0 transpositions.

@Vercassivelaunos So does that mean my proof is sound?

@Math55: Your proof is absolutely correct and complete. With this correction $\sigma\neq e$.

It is also the product of an arbitrary non-negative even number of transpositions, @Vercassivelaunos.

Intuitively , it is clear that we can achieve every permutation with at most $n-1$ transpositions : If the first element is not the desired one, swap the desired element with the current first one. Repeat this for the second, third , and so on until the $n-1$ th element. If all numbers in the positions $1$ to $n-1$ are as desired, the last number must be correct as well.

To be honest, I don't really get your proof. What is $n_1$? And are you saying that any permutation is just a cycle?

@Shaun: it is, but that doesn't help in the case of $n\in\{1,2\}$

@Vercassivelaunos Sorry it should have been $n-1$. Just edited it

You can always multiply an element by the identity, too.

Well, $$e=(12)(12)(12)(12).$$

@Shaun For the given problem , this does not matter anyway , since the claim is only that we need AT MOST $n-1$ transpositions. "Exactly $n-1$" would be wrong anyway in general since the number of transpositions has a parity fixed by the permutation.

@Math55: I recommend looking at here

That's not what I'm saying, @Peter. Besides, if an element had $n-1$ transpositions, then just multiply by $(12)(12)$.

The problem is, @Peter, that the OP needed to specify that the transportations need to be disjoint.

I disagree with this being a duplicate of the linked question. While the answer to this question follows from those in the "duplicate", this question is more elementary, and can have helpful answers which are easier to grasp than the answers to the duplicate.