How to prove that $sign(\alpha \beta) = sign(\alpha)sign(\beta)$? Claim : How to prove that $\text{sign}(\alpha \beta) = \text{sign}(\alpha)\text{sign}(\beta)$, where $\alpha,\beta \in S_n$?


Where $\text{sign}(\alpha) = 1$ if number of transpositions in the representation in $\alpha$ is even.

and it will be -1 otherwise.

Case 1: $\alpha$ and $\beta$ both have even number of transpositions then it is easy to verify that above claim hold

Case 2 : if $\alpha$ has odd number of transpositions and $\beta$ has even number then $\text{sign}(\alpha\beta)$ will be -1 and 
$\text{sign}(\alpha) \text{sign}(\beta) $ will also be negative.

Then case 3 and case 4 and we are done. Is there any better proof for this problem?
Define $f:S_n\to GL_n(\mathbb{C})$ by $f(g)e_i = e_{g(i)}$ for the standard basis $e_1, \dots, e_n$ of $\mathbb{C}^n$. This map is clearly a homomorphism. Then $\det f:S_n\to \{\pm 1\}$ has $f(g) = -1$ if $g$ is a transposition (for example, use the invariance of $\det$ under a change of basis to reduce the case of $g = (12)$ and compute it directly), so $f(g) = \operatorname{sign}(g)$ for any $g$.
Given the definition and the observation that the number of transpositions in (some representative of) $(\alpha + \beta)$ is the sum of that in $\alpha$ and in $\beta$, this is equivalent to proving that odd + odd is even, etc. You can phrase that equivalently as the addition being well-defined $\mod 2$ and so on, if that counts as a 'better' proof.