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For finite sets $A_1,\dots ,A_n$, the inclusion-exclusion principle statesTo prove it, fix an element belonging to exactly $k$ of the sets. If $k=0$, it contributes zero to both sides. If $k\ge 1$, its total weight on the right isby the binomial theorem. Summing these pointwise contributions proves the formula.

Let the universe be the $n!$ permutations of $\{1,\dots ,n\}$, and let $A_j$ be the set of permutations fixing $j$. A derangement belongs to none of the $A_j$. If $k$ specified points must be fixed, the remaining $n-k$ points can be permuted in $(n-k)!$ ways, so every such $k$-fold intersection has that size. There are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ choices of the specified fixed points. Applying inclusion-exclusion to the complement of their union gives the derangement counting formulaThe $k=0$ term is the whole universe. Since the exponential power series converges absolutely at $-1$,The alternating-series remainder additionally gives $\left|f(n)/n!-e^{-1}\right|\le 1/(n+1)!$. Thus the fraction is the probability of no fixed points in a uniformly chosen permutation and approaches $1/e$.