Calculation of a Permutation's Sign

Let $\pi\in S_n$ be a permutation taken from the symmetrical group then it is easy to see that it can be expressed as the product of cyclic permutations $c_i\in S_n$ as follows: $\pi = \prod_{i=1}^k c_i$. And each cycle $c_i$ can be expressed as a product of transpositions $t^i_j:=(c_i^{j-1}(1) ~ c_i^j(1))\in S_n$ for $j=1,2,...,|c_i|-1$ $$c_i=(1~c_i(1)~c_i^2(1)~... ~c_i^{|c_i|-1}(1))=(1~ c_i(1))(c_i(1)~ c_i^2(i))...(c_i^{|c_i|-2}(1)~ c_i^{|c_i|-1}(1))$$ Here the expression $|c_i|$ is the order (cardinality) of the group $<c_i>$. Also note that $c_i^{|c_i|}=\mathrm{id}$ where $\mathrm{id}$ is the identity of $S_n$.

A transposition $(i~j)$ has $2(j-i)-1$ inversions. The inversions are the pairs $(i,k)$ for $k=i+1,i+2,...,j$ and the pairs $(k',j)$ for $k'=i+1,i+2,...,j-1$ so it all comes down to basic counting. Note that every other element besides $i,j$ are fix points. Which means the sign of a (simple) transposition is $-1$. 

The number of inversions of the identity is zero. Are there any other permutations besides the identity with no inversions? Let $\alpha\in S_n$ have zero inversions but $\alpha\neq\mathrm{id}$. Since this permutation doesn't have any inversions the chain $\alpha(1) < \alpha(2) < ... < \alpha(n)$ must be valid because $\alpha(i)<\alpha(j) \forall i<j$ is given by the definition of this permutation. In other words $\alpha(1)$ must be thre smallest element of the sequence $(1,2,...,n)$ which is $1$. $\alpha(2)$ the second smallest element which is $2$ and so on $\Rightarrow \alpha=\mathrm{id}$. There are no permutations besides the identity with no inversions.

$\pi$ is a permutation which is not the identity, so it has inversions. Can we find a simple transposition $\tau$ such that $\tau\pi$ has less inversions than $\pi$? If there would be such a simple transposition then there also exists a an inverse image $i$ with $\pi(i) > \pi(i+1)$. And this $i$ has to exist because otherwise we would have $\pi(i) < \pi(i+1)$ for all $i=1,2,...,n-1$ which is only true for the idenity - contradiction to the assumptions $\pi\neq \mathrm{id}$. Therefore we can find simple transpositions to non trivial permutations to reduce their inversion count by one. 

A transposition $(i~j)$ as seen above has $2(j-i)-1$ inversions. Therefore it takes $2(j-i)-1$ simple transpositions $\tau_h$ such that $\left(\prod_{h=1}^{2(j-i)-1}\tau_h\right) (i ~ j) = \mathrm{id}$ and such transpositions do exist. This also means that we can write every permutation as a product of simple transpositions.

The sign mapping $\mathrm{sgn}:S_n\rightarrow \{-1,1\}$ is a group homomorphism. To show that we first need to realize that the sign of a permutation $\pi$ changes when we multiply the permutation with a simple transposition $\tau$ since we either add or remove a single inversion. $$\mathrm{sgn}(\pi\tau) = \mathrm{sgn}(\tau\pi)=-\mathrm{sgn}(\pi)=\mathrm{sgn}(\tau)\mathrm{sgn}(\pi)$$ Especially we get for simple transpositions $\tau_1, \tau_2$ $$\mathrm{sgn}(\tau_2\tau_1)=\mathrm{sgn}(\tau_2)\mathrm{sgn}(\tau_1)$$

Let $\pi\pi'\in S_n$ be permutations which are not the identity. There exist $N, N'$ simple transpositions $\tau_i, \tau'_i$ such that $\pi=\prod_{i=1}^N\tau_i$ and $\pi'=\prod_{i=1}^{N'}\tau'_i$. This yields $$\mathrm{sgn}(\pi\pi')=\mathrm{sgn}\left(\left(\prod_{i=1}^N\tau_i\right)\left( \prod_{i=1}^{N'}\tau'_i\right)\right)=  \mathrm{sgn}\left(\prod_{i=1}^N\tau_i\right)  \mathrm{sgn}\left( \prod_{i=1}^{N'}\tau'_i\right) =\mathrm{sgn}(\pi)\mathrm{sgn}(\pi')$$  

Let's take a closer look at the permutation $\pi$ as it is defined directly above. We come to the formula $$\mathrm{sgn}(\pi)=\prod_{i=1}^N\mathrm{sgn}(\tau_i)=(-1)^N$$ This tells us that if $pi$ is even than $N$ must even as well. The same idea applies to $\pi$ being an odd permutation.

Back to cyclic permutations. Every transposition is odd and therefore every odd cyclic permutation must be a product of an odd number of (simple) transpositions. The cycles $c_i$ which we defined at the beginning are odd if $|c_i|$ is even because we can express $c_i$ with the product of $|c_i|-1$ transpositions. The sign of $pi$ is calculated as $$\mathrm{sgn}(\pi)=\prod_{i=1}^k\mathrm{sgn}(c_i)$$ The signs on the right side of the equation which equal $1$ can be discarded in the process of the calculation (neutral element). So we can come to the conclusion:

In Order to calculate the sign of a permutation $\pi$ we express it as a product of cyclic permutations. Then we count the even cycles, let us assume there are $K$ such cycles, and calculate the sign $$\mathrm{sgn}(\pi)=(-1)^K$$

This method is straight forward and not as error prone as to count each single inversion manually.