Pick $\sigma_1$. That determines that there will be $\sigma_1-1\in\{0,\dotsc,n-1\}$ inversions involving $\sigma_1$, no matter what other values you will pick. The options you have for inversions involving only the remaining values don't depend on the value of $\sigma_1$ you picked; you can imagine the remaining values "compressed" into the range $\{2,\dotsc,n\}$ by moving the values below $\sigma_1$ up by one to fill the gap; that doesn't change the ordering of the values. So the options remaining are the ones for a permutation of $n-1$ objects. Thus the formula follows by induction, since the options $0,\dotsc,n-1$ for $\sigma_1-1$ add a factor $1+x+\dotso+x^{n-1}$ to the generating function for $S_{n-1}$.