Derangement recurrence

Fix element $n$. In a derangement, $n$ goes to position $k\ne n$.

Case 1: $\sigma(k)=n$. Then $\sigma$ on the remaining $n-2$ elements is a derangement of those elements: $D_{n-2}$.

Case 2: $\sigma(k)\ne n$. Then the restriction to $\{1,\ldots,n-1\}\setminus\{k\}$ with $\sigma(j)\ne j$ and $\sigma(j)\ne k$ is a derangement of $n-1$ elements: $D_{n-1}$.

Total. $(n-1)(D_{n-1}+D_{n-2})$.

Conclusion. $D_n=(n-1)(D_{n-1}+D_{n-2})$.