Recall that aderangementof a set $\displaystyle S$ is a permutation $\displaystyle \pi$ of $\displaystyle S$ having no fixed points, i.e. $\displaystyle \pi(s)\neq s \ \ \forall s \in S$. If the number of derangements of $\displaystyle \{1,\dots,n\}$ is $\displaystyle !n$, show that

$\displaystyle !n = \sum_{j=0}^n{n \choose j}(-1)^{n-j} j!$.

Hint :

Spoiler: