Proof on a property of Stirling Numbers of the first kind How should i proceed to prove that the sum of every odd stirling number on a row is n!/2?
$$\sum\limits_{k=1|k=odd}^n s(n,k)=\frac{n!}{2}$$
 A: HINT:
$s(n, k)$ is the number of permutations of $1, 2, ..., n$ that can be expressed as a product of $k$ cycles with disjoint orbits. Therefore, $n! = s(n, 0) + ... + s(n, n)$.
Show that $\sum\limits_{k=1|k \equiv 0 \mod 2}^n s(n,k)=\sum\limits_{k=1|k \equiv 1 \mod 2}^n s(n,k)$
And you're done. For that, define a bijection between two sets whose cardinalities are $\sum\limits_{k=1|k \equiv 0 \mod 2}^n s(n,k)$ and $\sum\limits_{k=1|k \equiv 1 \mod 2}^n s(n,k)$.
A: Recall that permutations by cycles are the following species:
$$\mathfrak{P}(\mathfrak{C}(\mathcal{Z})).$$
Therefore the bivariate generating function of permutations by the number of cycles is
$$\exp\left(u\log\frac{1}{1-z}\right).$$
It follows that the generating function of permutations with an odd number of cycles is
$$\frac{1}{2}\exp\left(+1\times\log\frac{1}{1-z}\right)
-\frac{1}{2}\exp\left(-1\times\log\frac{1}{1-z}\right).$$
This simplifies to
$$\frac{1}{2}\frac{1}{1-z}-\frac{1}{2}(1-z).$$
Extracting coefficients we get that for $n\ge 2$
$$n! [z^n]\left(\frac{1}{2}\frac{1}{1-z}-\frac{1}{2}(1-z)\right)
=\frac{1}{2}n!$$
We get for $n=1$ the result $1/2+1/2=1$ and for $n=0$, zero. 
