Proof that $\left.\frac{d^{2n}}{ds^{2n}}\right|_{s=0} e^{s^2/2}=(2n-1)!!$ In the very first example (page 4) of Matrix Model Combinatorics: Applications to Folding and Coloring, the author claims that the nonzero contributions to $\left.\frac{d}{ds}\right|_{s=0} e^{s^2/2}$ are in bijection with perfect matchings on $\{1,\dots,2n\}$.
Somehow this has something to do with 'taking the derivative in pairs', but I don't see it. I wrote out the first few derivatives for this expression: $e^{s^2/2}, se^{s^2/2}, e^{s^2/2} + s^2 e^{s^2/2}, \dots,$ but it didn't help.
I can prove this fact by induction but I'd like to understand what the author meant.
In particular I'd like to understand what the correspondence between 'nonzero contributions' and 'perfect matchings' is (Figure 1. in the paper).
 A: After reading ahead a little bit, I think I've figured out what the author meant. 
The key is the following: every time we get a factor of $s$ from differentiating, we need to keep track of when it appeared.
So the first three derivatives are (writing $f$ for $e^{s^2/2}$ so that $\frac{d}{ds} f = sf$):


*

*$s_1 f$

*$f+ s_1 s_2 f$

*$s_3 f + s_2 f +  s_1 f +  s_1 s_2 s_3 f$


Every time we differentiate a term of the form $s^k f$, we have a choice of either adding a factor $s_i$, or removing one of the pre-factors $s_j$. 
For a term to have non-zero contribution, there must be no pre-factors of any $s_i$. This means that any $s_i$ that appears must be differentiated at some later step $j$. The matching corresponding to a contributing term can be found by tracing through the `differentiation history' of the term and writing down the collection of pairs $(i,j)$.
This is a complete matching because if we trace through the `history' of a contributing term, at each stage, either a pre-factor $s$ was added, or a pre-factor was removed.
