Sign of composition of transpositions Let $\sigma \in S_n$.
Definition: Suppose that $\text{sign}\sigma=(-1)^N$, where $N$ - number of inversions in permutation $\sigma$.
Suppose that $\tau_1$ and $\tau_2$ transpositions. How to prove that $\text {sign}(\tau_1\circ \tau_2)=\text {sign}\tau_1\cdot \text {sign}\tau_2?$
I tried to prove it but any results. 
Can anyone show the rigorous proof please.

During a long drive this evening I realized my other answer, involving determinants, was just a long trip around Robin Hood's Barn. Here's the short proof. 
Suppose that 
$\tau_1$ is a composition of $k$ transpositions $X_1, \ldots, X_k$, 
$$
\tau_1 = X_k X_{k-1} \cdots X_2 X_1,
$$
and that
$\tau_2$ is a composition of $p$ transpositions $Y_1, \ldots, Y_p$. Then $\text{sign}(\tau_1) = (-1)^k$, and similarly  $\text{sign}(\tau_2) = (-1)^p$. 
To compute the sign of $\tau_2 \circ \tau_1$, observe that $\tau_2 \circ \tau_1$ can be written as 
$$
\tau_2 \circ \tau_1 = Y_p Y_{p-1} \cdots Y_2 Y_1 X_k X_{k-1} \cdots X_2 X_1,
$$
so that its sign is $(-1)^{p+k}$. And since
$$
(-1)^{p+k} = (-1)^p (-1)^k
$$
we get the desired conclusion. 
There's an important question here: why can a single permutation not be written as a product of $k$ permutations and of $b$ permutations, where exactly one of $b$ and $k$ is odd? I.e., why is the sign well-defined in the first place? I'm assuming that question has already been covered in your text...
