Proof/Reasoning why the sgn function which counts inversions has the following property? $\mathrm{sgn}(\pi\circ\sigma)=\mathrm{sgn}(\pi)\cdot \mathrm{sgn}(\sigma)$
I am familiar with how to count inversions and any insight for why this formula holds true would be very helpful.
 A: Suppose that $\langle i,j\rangle$ is an inversion of $\pi\circ\sigma$. Then $i<j$, and $\pi\big(\sigma(i)\big)>\pi\big(\sigma(j)\big)$. 


*

*If $\langle i,j\rangle$ is not an inversion of $\sigma$, we have $\sigma(i)<\sigma(j)$, so $\langle\sigma(i),\sigma(j)\rangle$ must be an inversion of $\pi$.  

*If $\langle i,j\rangle$ is an inversion of $\sigma$, we have $\sigma(i)>\sigma(j)$, so $\langle\sigma(j),\sigma(i)\rangle$ is not an inversion of $\pi$.


Similarly, if $\langle i,j\rangle$ is not an inversion of $\pi\circ\sigma$, then either 


*

*$\langle i,j\rangle$ is not an inversion of $\sigma$, and $\langle\sigma(i),\sigma(j)\rangle$ is not an inversion of $\pi$, or  

*$\langle i,j\rangle$ is an inversion of $\sigma$, and $\langle\sigma(j),\sigma(i)\rangle$ is an inversion of $\pi$.


Now imagine a three-column listing of pairs. The first column is simply a list of all pairs $\langle i,j\rangle$ with $1\le i<j\le n$. The second and third columns will be filled with $1$s and $-1$s. Specifically, for the row headed by $\langle i,j\rangle$, the second column contains $-1$ if $\langle i,j\rangle$ is an inversion of $\sigma$ and $1$ if it is not. The third column contains $-1$ if $\langle\sigma(i),\sigma(j)\rangle$ (or $\langle\sigma(j),\sigma(i)\rangle$, whichever is appropriate) is an inversion of $\pi$ and a $1$ if it is not. 
The product of the numbers in the second column is clearly $\operatorname{sgn}(\sigma)$, and the fact that $\sigma$ is a bijection on $\{1,\ldots,n\}$ ensures that the product of the numbers in the third column is $\operatorname{sgn}(\pi)$. The product of these two column products is therefore $\operatorname{sgn}(\pi)\operatorname{sgn}(\sigma)$.
Now make a fourth column whose entries are the products of the entries in the second and third columns. The four bullet points above ensure that the entry in the fourth column of row $\langle i,j\rangle$ is $-1$ if $\langle i,j\rangle$ is an inversion of $\pi\circ\sigma$ and is $1$ if it not. Thus, the product of the numbers in the fourth column is $\operatorname{sgn}(\pi\circ\sigma)$. Finally, this product is clearly equal to the product of all of the numbers in the second and third columns, which we just saw is $\operatorname{sgn}(\pi)\operatorname{sgn}(\sigma)$.
A: Consider permutation group on $4$ letters only,  since you will find arguments simple. One can then easily see it for general $n$. 
Consider the polynomial $$\Phi(x_1,x_2,x_3,x_4)=(x_1-x_2)(x_1-x_3)(x_1-x_4)(x_2-x_3)(x_2-x_4)(x_3-x_4).$$
Let $\sigma.\Phi$ denote the polynomial obtained by permuting the indices in $\Phi$ according to $\sigma$. Thus, for $\sigma=(12)$, we get 
$$\sigma.\Phi(x_1,x_2,x_3,x_4)=(x_2-x_1)(x_2-x_3)(x_2-x_4)(x_1-x_3)(x_1-x_4)(x_3-x_4).$$
Notice that the factors in $\sigma.\Phi$ and $\Phi$ are same except for change of sign. Also notice that $\sigma$ is inverting factors $x_1-x_2$ only (other factors appear as same).
Thus, the number of inversions by $\sigma$ is just $1$. 
Exercise: Think for number of inversions for $(12)(34)$ and $(123)$. 
One simple way to talk about sign of $\sigma$ is that it can be defined as the ratio $\frac{\sigma.\Phi}{\Phi}$. Then, 
$$\mathrm{sign}(\sigma\tau)=\frac{\sigma\tau.\Phi(x_1,\cdots,x_n)}{\Phi(x_1,\cdots,x_n)}=\frac{\Phi(x_{\sigma\tau(1)},\cdots,x_{\sigma\tau(n)})}{\Phi(x_1,\cdots,x_n)}.$$
Multiply and divide by $\Phi(x_{\tau(1)},\cdots,x_{\tau(n)})$:
$$\mathrm{sign}(\sigma\tau)=\frac{\Phi(x_{\sigma\tau(1)},\cdots,x_{\sigma\tau(n)})}{\Phi(x_{\tau(1)},\cdots,x_{\tau(n)})}.\frac{\Phi(x_{\tau(1)},\cdots,x_{\tau(n)})}{\Phi(x_1,\cdots,x_n)}.$$
Denote by $y_i$ the variable $x_{\tau(i)}$. Then you can see that
$$\mathrm{sign}(\sigma\tau)=\frac{\sigma.\Phi(y_1,\cdots,y_n)}{\Phi(y_1,\cdots,y_n)}.\frac{\tau.\Phi(x_{1},\cdots,x_{n})}{\Phi(x_1,\cdots,x_n)}=\mathrm{sign}(\sigma)\mathrm{sign}(\tau).$$
