Let \(n\) be a positive integer, and let \(D(n)\) be the set of arrangements of \(\{1, \ldots, n\}\) for which no number is in its corresponding position. (For example, if \(n=4\) then the arrangement \(4,2,3,1\) is not in \(D(4)\) as the number 2 is in position 2; but the arrangement 4,3,2,1 is in \(D(4)\).) Use the Inclusion-Exclusion Principle to prove that

\[ |D(n)|=n!\left(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\cdots+(-1)^{n} \frac{1}{n!}\right) \]