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What is wrong with the following inductive “proof” that D(n) = (n−1)! for all n ≥ 2? Can you find a false step in it?

For n=2, the formula holds, so assume n ≥ 3. Let π be a permutation of {1,2,...,n−1} with no fixed point. We want to extend it to a permutation π′ of {1,2,...,n} with no fixed point. We choose a number i ∈ {1, 2, . . . , n − 1}, and we define π′(n) = π(i), π′(i) = n, and π′(j) = π(j) for j ̸= i,n. This defines a permutation of {1,2,...,n}, and it is easy to check that it has no fixed point. For each of the D(n − 1) = (n − 2)! possible choices of π, the index i can be chosen in n − 1 ways. Therefore, D(n) = (n − 2)! · (n − 1) = (n − 1)!.

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    Have you compared to the actual number of derangements? Since it's presumably not the right formula, it must over- or under-estimate the actual number at some point; have you tried seeing which occurs when the two differ? It happens for a small value of $n$. – 2017-02-07

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The argument only covers permutations where $n$ is in $2-$cycle. For example, it does not cover the permutation $(1,2,3,\dots n)$, which is still a derangement.

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    Yes, since when it goes over 2, it will be 3->(2)!,4->(3!) .....n -> (n-1)! if there is 1 fixed point. – 2017-02-07