 Solution

In the set of all $n!$ permutations, let $A_i$ consist of those fixing $i$. For a prescribed collection of $k$ fixed points, its intersection has $(n-k)!$ members. Applying the inclusion-exclusion principle to the complement of ${\bigcup }_i{A}_i$ gives the number of derangements:The convergent power series for the exponential function at $-1$ therefore gives $\boxed{d_n/n!⟶e^{-1}}$.