EDIT:
PROOF BY INDUCTION: If there are no inversions then we have an identity permutation, so the base case holds.
INDUCTIVE STEP : Lets fix some $r \in \mathbb{Z}^{+}$ and take any $\pi \in S_n$. If $\pi$ is the composition of inversions then $sgn(\pi) = (-1)^r$.
Fix some $\pi \in S_n$ then $\pi$ can be written as a composition of r+1 inversions. Lets say it can be written as follows :
$$\pi = (a_0 a_0')\circ(a_1 a_1')\circ \dots \circ(a_r a_r')$$
And lets say
$$\pi' = (a_1 a_1')\circ \dots \circ(a_r a_r')$$
And the induction hypothesis satisfies $sgn(\pi') = (-1)^r$
But $\pi = (a_0 a_0') \circ \pi'$
And we know as a fact that $sgn((i j) \circ \pi) = -sgn(\pi)$. Which implies $$sgn(\pi) = -sgn(\pi')$$ $$sgn(\pi) = -(-1)^r = (-1)^{r+1} \quad \square$$