How can one derive from the following equality the number of even n-permutations (i.e. the number of n-permutations with even number of inversions)?
$$\sum_\limits{k=0}^n c(n,k)x^k = x(x+1)(x+2)\dots (x+n-1) $$
where $c(n,k)$ is Stirling's number of the first kind.
Note: number of even n-permutations is $\frac{n!}{2}$
A permutation of $n$ objects with $k$ cycles is even if $n-k$ is even and odd if $n-k$ is odd. So the number of even permutations is $$c(n,n)+c(n,n-2)+\cdots$$ and of odd permutations is $$c(n,n-1)+c(n,n-2)+\cdots.$$ The difference of these is $$c(n,n)-c(n,n-1)+c(n,n-2)-c(n,n-3)+\cdots =(-1)(-1+1)(-1+2)\cdots=0$$ for $n\ge2$.