Note that if:
$\alpha = \sigma_1\sigma_2\cdots\sigma_k$
that:
$\beta\alpha\beta^{-1} = (\beta\sigma_1\beta^{-1})(\beta\sigma_2\beta^{-1})\cdots(\beta\sigma_k\beta^{-1})$.
Now if each $\sigma_i$ for $i = 1,\dots,k$ is a transposition, we will show each $\beta\sigma_i\beta^{-1}$ is likewise a transposition.
Suppose $\sigma_i = (a\ b)$. Let us denote:
$c = \beta(a)$ and $d = \beta(b)$, so that $\beta^{-1}(c) = a$, and $\beta^{-1}(d) = b$.
If $m \not\in \{c,d\}$, then $\beta^{-1}(m) \not\in \{a,b\}$ (because $\beta,\beta^{-1}$ are bijective), and thus:
$\sigma_i\beta^{-1}(m) = \sigma_i(\beta^{-1}(m)) = \beta^{-1}(m)$, since $\sigma_i$ only affects $a$ and $b$.
Therefore, $\beta\sigma_i\beta^{-1}(m) = \beta(\sigma_i(\beta^{-1}(m))) = \beta(\beta^{-1}(m)) = m$.
If $m = c$, we have: $\beta\sigma_i\beta^{-1}(c) = \beta(\sigma_i(\beta^{-1}(c))) = \beta(\sigma_i(a)) = \beta(b) = d$,
and if $m = d$, we have: $\beta\sigma_i\beta^{-1}(d) = \beta(\sigma_i(\beta^{-1}(d))) = \beta(\sigma_i(b)) = \beta(a) = c$.
So, $\beta\sigma_i\beta^{-1} = (c\ d)$.
Thus if $\alpha$ is a product of $k$ transpositions, $\beta\alpha\beta^{-1}$ is likewise, whether $k$ be even or odd.
In your proof, you are assuming that a permutation cannot be both even and odd (for otherwise, you could express B as a product of an even number of transpositions, and B^(-1) as a product of an odd number of transpositions, and then your proof falls flat). To prove that a permutation cannot be BOTH even and odd, you need the fact that identity can only be expressed as a product of an even number of permutations. So you do need an alternative approach to proving this result. That's why most proofs are lengthy.
Best Answer
If $\alpha = \sigma_1\sigma_2\dots \sigma_{2k}$, where $\sigma_i$ is transposition then $\alpha^{-1} = \sigma_{2k}\dots \sigma_2\sigma_1$.
Similarly with the case of odd permutation.