Non-enumerative proof that there are many derangements? Recall that a derangement is a permutation $\pi: \{1,\ldots,n\} \to \{1,\ldots,n\}$ with no fixed points: $\pi(j) \neq j$ for all $j$.  A classical application of the inclusion-exclusion principle tells us that out of all the $n!$ permutations, a proportion $1/e + o(1)$ of them will be derangements.  Indeed, by computing moments (or factorial moments) or using generating function methods, one can establish the stronger result that the number of fixed points in a random permutation is asymptotically distributed according to a Poisson process of intensity 1.
In particular, we have:
Corollary: the proportion of permutations that are derangements is bounded away from zero in the limit $n \to \infty$.
My (somewhat vague) question is whether there is a "non-enumerative" proof of this corollary that does not rely so much on exact combinatorial formulae.  For instance, a proof using the Lovasz Local Lemma would qualify, although after playing with that lemma for a while I concluded that there was not quite enough independence in the problem to make that lemma useful for this problem.
Ideally, the non-enumerative proof should have a robust, "analytic" nature to it, so that it would be applicable to other situations in which one wants to lower bound the probability that a large number of weakly correlated, individually unlikely events do not happen (much in the spirit of the local lemma).  My original motivation, actually, was to find a non-enumerative proof of a strengthening of the above corollary, namely that given $l$ permutations $\pi_1,\ldots,\pi_l: \{1,\ldots,n\} \to \{1,\ldots,n\}$ chosen uniformly and independently at random, where $l$ is fixed and $n$ is large, the probability 
that these $l$ permutations form a $2l$-regular graph is bounded away from zero in the limit $n \to \infty$.  There is a standard argument (which I found in Bollobas's book) that establishes this fact by the moment method (basically, showing that the number of repeated edges or loops is distributed according to a Poisson process), but I consider this an enumerative proof as it requires a precise computation of the main term in the moment expansion.  
 A: *

*The mean number of fixed points is 1.  This is very elementary.

*Consider the operation of rotating three values around: $p(i)\to p(j)\to p(k)\to p(i)$. Given a permutation with no fixed points, there are $n^2-O(n)$ rotations that create from it a permutation with exactly one fixed point. Given a permutation with exactly one fixed point, there are $n^2-O(n)$ rotations that create from it a permutation with no fixed points.

*From (2), the numbers of permutations with 0 and 1 fixed points are the same to $O(1/n)$ relative error.  Combining this with (1) shows that the fraction of permutations with no fixed points is at least $\frac13-O(1/n)$.
This is the switching method and it can be used in extremely many circumstances. By continuing to increase the number of fixed points by further switchings, step 1 could be avoided.
Incidentally, congratulations to Terry and Jean Bourgain.
A: There is a very elegant   approach to derangements

D.M. Jackson, Laguerre polynomials and
  derangements, Math. Proc. Camb. Phil.
  Soc., 80 (1976), 213–214.

I am not sure    it fits   precisely your criteria, but I found   Jackson's approach very useful for  various asymptotic estimates.
A: I'm not sure this qualifies, but here goes.
It doesn't take any fancy enumerations to show that the average number of fixed points is 1:  Among the $n!$ permutations of $n$ objects, each object is fixed by $(n-1)!$ permutations, for a total of $n!$ fixed points.  So if the probability of having no fixed points tends to 0 (for some subsequence of $n$'s), then the probability of having more than one fixed point must also tend to 0.  This seems like it should be easily disprovable nonsense.
Added 1/20/12  Brendan McKay's "switching method" answer nails the nonsense.  Here's a slight variation on his argument.
Assume you're at a large $n$ where, say 99 percent (or more) of the permutations have exactly one fixed point.  (That was the thrust of my intial posting:  If the fraction of permutations with no fixed points is close to 0, then the fraction with exactly one fixed point must be close to 1.)  Then there's at least one object that's fixed by at least $.99(n-1)!$ permutations having no other fixed points.  If you take each of these permutations and switch the fixed object with each of the other $n-1$ objects, you've created $.99(n-1)!(n-1) = .99n!(1-1/n)$ distinct permutations with no fixed points.  Since $n$ is large, this is almost $.99n!$ itself, so certainly greater than $.5n!$.  But $.99+.5 > 1$, which gives us more than $n!$ permutations on $n$ objects.
A: L. Lu and L. A. Szekely have successfully applied the Lopsided (i.e. Negative Dependency Graph) Lovasz Local Lemma to this problem.
A negative dependency graph is as a dependency graph except that independence is replaced with the inequality
$$
\Pr\left(A_k \middle\vert \bigwedge_{i \in S} A_i\right) \leq \Pr(A_k)
$$
for any fixed event $A_k$ and collection $S$ of non-neighbors of $A_k$. Erdos and Spencer showed the Lovasz Local Lemma holds with negative dependency graphs in place of dependency graphs.
Let $\Omega$ be the probability space of all perfect matchings of $K_{n,n}$ equipped with the uniform distribution. For a partial matching $M$ of $K_{n,n}$, define the event
$$
A_M = \{M^\prime \in \Omega \mid M \subseteq M^\prime\}
$$
(all perfect matching that contain the partial matching $M$).
Given a collection $\mathcal{M}$ of partial matchings of $K_{n,n}$, construct a graph with vertex set $\{A_M \mid M \in \mathcal{M}\}$ and set two matchings adjacent if their union is not again a matching. L. Lu and L. A. Szekely showed this graph is a negative dependency graph.
Finally we can address the problem at hand. Let the partite sets of $K_{n,n}$ be $\{1, \dots, n\}$ and $\{1^\prime, \dots, n^\prime\}$. For $1 \leq i \leq n$, let $M_i$ be the one-edge matching $ii^\prime$. Viewing perfect matchings of $K_{n,n}$ as permutations of an $n$-element set, the event $\bigwedge_{i=1}^n \overline{A_{M_i}}$ contains precisely those permutations not having a fixed point. Choosing $x_i = \frac{1}{n}$ for the purposes of the Lopsided Lovasz Local Lemma, we get
$$
\Pr\left( \bigwedge_{i=1}^n \overline{A_{M_i}} \right) \geq \left(1 - \frac{1}{n}\right)^n,
$$
which converges to $\frac{1}{e}$ as $n \rightarrow \infty$.
A: The fraction $f(n)$ of permutations of $n$ letters that are derangements satisfies the recurrence $f(n) - f(n-1) = \frac{-1}{n} (f(n-1) - f(n-2))$ with $f(0)=1$ and $f(1)=0$.
It is easy to see from this that $f(2k) > f(2k+1)>f(2k-1)$, and in particular $f(n) \ge f(3) > 0$ for $n \ge 2$.
A: The number of derangements of an $n$-set is the number of perfect matchings
in the bipartite complement of $n$ disjoint copies of $K_2$. The rook polynomial
of $nK_2$ is $(x-1)^2$ and so the number of matchings in the complement is equal to
$$
 \int_0^\infty (x-1)^n e^{-x}\ dx.
$$
(I am not sure if this is along the lines you want, but it is a nice formula for asymptotic
purposes.) This approach is used in anger in C. D. Godsil and B. D. McKay, Asymptotic enumeration of Latin rectangles, J. Combinatorial Theory, Ser. B, 48 (1990) 19-44. (Also on Brendan's web page.)
