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Not certain whether my proof is right, would appreciate it if I could get some feedback on it. Also, the epsilon here is the sign function of the permutation so $epsilon=sgn$

Proof:

Since the mapping $ , , epsilon:S_nrightarrowpm1$ is a group homomorphism, I'll use the fact that
beginequation
epsilon(sigmatau)=epsilon(sigma)cdotepsilon(tau)
endequation
And let $tau=sigma^-1$ which will give us $epsilon(sigmasigma^-1)=epsilon(e)$ with $e$ the identity permutation. And since the identity permutation has sign $1$ we have that $epsilon(sigma)cdotepsilon(sigma^-1)=1$.

Now since $epsilon(sigma)=(-1)^textnumber of inversions of , sigma$ and $epsilon(sigma^-1)=(-1)^textnumber of inversions of , sigma^-1$ let $n$ and $m$ denote those powers respectively (to avoid cumbersome notation) we then have that
beginalign
&(-1)^ncdot(-1)^m=1\
&(-1)^n+m=(-1)^2\
&n=m-2
endalign
And since $m-2$ doesn't alter the sign of $(-1)$ (there's a better way of saying this) we have that the number of both inversions is the same and hence the sign of both permutations is also the same.

How does this look?

You can't conclude that $n = m - 2$, only that $n+m$ is even, but that's enough to finish. (It's also just not true: $(12)^-1 = (12)$, so in this case $n=m$, not $n=m-2$.)

You can also stop at $epsilon(sigma) cdot epsilon(sigma^-1) = 1$. This already tells you that either $epsilon(sigma)=epsilon(sigma^-1) = 1$ or $epsilon(sigma)=epsilon(sigma^-1) = -1$.

When you say $n=m-2$, that should be $nequiv m-2mod 2$. The value of $epsilon(sigma)$ doesn't tell you what that exponent is, only that it's even or odd.

In fact, nothing about this depends on that "number of inversions" formula. It's entirely a consequence of $epsilon$ being a homomorphism to the two-element group.

The way I would phrase it? $epsilon(sigma^-1)=(epsilon(sigma))^-1$ since $epsilon$ is a homomorphism. Then, in the two-element group $1,-1$, every element is its own inverse, so $(epsilon(sigma))^-1=epsilon(sigma)$. Done.

You cannot conclude from $(-1)^n+m=(-1)^2$ that $n=m-2$. What you can conclude is that $n+m$ is even, and that is enough, because that means that $m$ and $n$ have the same parity.

There is a shorter proof: just note that assuming what you want is not true, then $epsilon(sigma) cdot epsilon(sigma^-1) = -1$.