Why is the Derangement Probability so Close to $\frac{1}{e}$? A derangement is a permutation $\sigma$ of $\{1,2,3,\dots,n\}$ such that $\sigma(i) \neq i$ for every $i$.  A common application of inclusion/exclusion in undergraduate combinatorics and probability classes is to compute the number of derangements, and in the process show that the probability a random permutation is a derangement approaches $\frac{1}{e}$ for large $n$.  
There's also a "standard" intuition for this probability, which goes roughly as follows:  Let $E_i$ be the event that $\sigma(i)=i$.  
1) For a given $i$, the probability of $E_i$ is exactly $\frac{1}{n}$.
2) If $n$ is large, than these events should be "nearly" independent ($E_i$ occurring means that $\sigma(i) \neq j$, making it a tiny bit more likely that $\sigma(j)=j$, but this shouldn't have much of an effect for large $n$), so we'd expect the probability none of the $E_i$ occur to be roughly $\left(1-\frac{1}{n}\right)^n$.
3) For large $n$, $\left(1-\frac{1}{n} \right)^n \approx \frac{1}{e}$.  
Now the last approximation alone already has an error proportional to $\frac{1}{n}$.  What's suprising then is that, after working through the inclusion/exclusion, you find that the probability is not just approximately $\frac{1}{e}$, but incredibly close -- the error is less than $\frac{1}{(n+1)!}$. 
Is there some alternative intuitive explanation for the $\frac{1}{e}$ asymptotic probability that gives a sense of why the convergence is so fast?

I'm not sure this is really better than the inclusion exclusion, but it is different:
Let $d_n$ be the number of derangements. Then we have
$$d_n = (n-1) (d_{n-1} + d_{n-2})$$
(see here). Let $p_n = d_n/n!$ be the probability of a derangement. Then
$$p_n = \frac{n-1}{n} p_{n-1} + \frac{1}{n} p_{n-2} \ \mbox{which implies} \ p_{n} - p_{n-1} = - \frac{p_{n-1} - p_{n-2}}{n} .$$
So $p_{n} - p_{n-1}$ decays factorially fast, and thus $p_n = \sum_{m=2}^n (p_m - p_{m-1})$ is factorially close to $\sum_{m=2}^{\infty} (p_m - p_{m-1}) = \lim_{n \to \infty} p_n$.
