How many 2m-permutations, consisting only of cycles of even length? How many 2m-permutations, consisting only of cycles of even length?
I have found this formula:
$$Q_2(n) =((2n − 1)!!)^2$$
but how it can be proven?
 A: Suppose we want to construct such a permutation $\pi$ of $1,\ldots, 2m$. 
Note that $\pi(1)$ must not be $1$, otherwise there would be a $1$-cycle.
So arbitrarily pick $x \in \{2,\ldots, 2m\}$ to be $\pi(1)$.  There are $2m-1$ possible choices.  Now if we take a permutation of $\{1,\ldots,2m\} \backslash \{1,x\}$ having only even cycles, expressed in disjoint cycle notation, and insert $1,x$  in one of those cycles following any element, we get a permutation of $\{1,\ldots, 2m\}$ with only even cycles.  Alternatively, we can insert $(1,x)$ as a cycle on its own.  All the allowable permutations of $\{1,\ldots,2m\}$ are obtained in this way, because if you take such a permutation where $x$ follows $1$ and erase $1$ and $x$ you obtain a permutation of $\{1,\ldots,2m\} \backslash \{1,x\}$ with only even cycles.
Thus we get the recursion
$$Q_2(m) = (2m-1) ((2m-2)+1) Q_2(m-1) = (2m-1)^2 Q_2(m-1)$$
with initial condition
$Q_2(1) = 1$.  The rest is induction.
A: The combinatorial species of permutations consisting only of even cycles is given by
$$\mathfrak{P}(\mathfrak{C}_{=2}(\mathcal{Z}) + \mathfrak{C}_{=4}(\mathcal{Z}) +
\mathfrak{C}_{=6}(\mathcal{Z})+ \mathfrak{C}_{=8}(\mathcal{Z}) + \cdots).$$
This translates to the generating function
$$G(z) = \exp\left(\frac{z^2}{2} + \frac{z^4}{4} + \frac{z^6}{6} + \frac{z^8}{8} + \cdots\right)$$
which is
$$\exp\left(\frac{1}{2}\left(\frac{z^2}{1} + \frac{z^4}{2} + \frac{z^6}{3} + \frac{z^8}{4} + \cdots\right)\right)$$
or
$$\exp\left(\frac{1}{2}\log\frac{1}{1-z^2}\right)
= \sqrt{\frac{1}{1-z^2}}.$$
By the Newton binomial we thus have the closed form for $n=2m$
$$(2m)! [z^{2m}] \sqrt{\frac{1}{1-z^2}}
= ((2m-1)!!)^2.$$
This is OEIS A001818.
Remark. The coefficient extraction can be performed by Lagrange inversion.
We have
$$[z^{2m}] G(z) 
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{2m+1}} \sqrt{\frac{1}{1-z^2}} \; dz.$$
This is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{(z^2)^{m+1}} \sqrt{\frac{1}{1-z^2}} \; z \; dz.$$
Put $1-z^2 = u^2 $ so that $- z \; dz = u\; du$ to get
$$- \frac{1}{2\pi i}
\int_{|u-1|=\epsilon} \frac{1}{(1-u^2)^{m+1}} \frac{1}{u} \times u \; du
\\ = - \frac{1}{2\pi i}
\int_{|u-1|=\epsilon} \frac{1}{(1-u)^{m+1}} \frac{1}{(1+u)^{m+1}}  \; du
\\ = (-1)^m \frac{1}{2\pi i}
\int_{|u-1|=\epsilon} \frac{1}{(u-1)^{m+1}} \frac{1}{(2+u-1)^{m+1}}  \; du
\\ = \frac{(-1)^m}{2^m} \frac{1}{2\pi i}
\int_{|u-1|=\epsilon} \frac{1}{(u-1)^{m+1}} \frac{1}{(1+(u-1)/2)^{m+1}}  \; du
\\ = \frac{(-1)^m}{2^m} \frac{1}{2\pi i}
\int_{|u-1|=\epsilon} \frac{1}{(u-1)^{m+1}} 
\sum_{q\ge 0} {q+m\choose m} (-1)^q \frac{(u-1)^q}{2^q} \; du
\\ = \frac{(-1)^m}{2^m} {2m\choose m} \frac{(-1)^m}{2^m}
= \frac{1}{2^{2m}} {2m\choose m}.$$
It follows that the answer is given by
$$(2m)! \times \frac{1}{2^{2m}} {2m\choose m}.$$
This implies that
$$\frac{Q_2(m)}{Q_2(m-1)}
= (2m)(2m-1) \frac{1}{2^2} \frac{(2m)(2m-1)}{m\times m} = (2m-1)^2$$
as pointed out by Robert Israel.
