Let $n ∈ \mathbb{N}$ be a natural number and $X$ a finite set with $n$ elements. Show that the number of permutations of $X$ such that no element stays in the same position is $$n!\sum_{k=0}^{n}\dfrac{(-1)^{k}}{k!}.$$
For instance, there are 6 = 3! permutations of 3 elements, but only 2 of them are permutations which fix no element. Similarly, there are 24 = 4! permutations of 4 elements, but only 9 which fix no element.
