# Reference for a combinatorial identity involving the number of derangements

Let $$c_n=n!\sum\limits_{k=0}^n (-1)^k \frac{1}{k!}$$ be the number of derangements of $$n$$ elements. The following combinatorial identity is coming up in my research:

$$\sum\limits_{j=1}^{n-2}c_{n-j}{n-2\choose j-1}=(n-2)!(n-2)$$

Is there a known reference or proof for this identity? Simulations do support this.

Let's try and count the number of permutations of $$\{1, \dotsc, n - 1\}$$ that do not fix $$1$$.
One way to count is to note that the number of permutations fixing $$1$$ is just the number of permutations of $$\{2, \dotsc, n - 1\}$$, so the answer is $$(n - 1)! - (n - 2)! = (n - 2)!(n - 2)$$. (Or you can argue that there are $$n - 2$$ choices for where to send $$1$$, then $$n - 2$$ choices for where to send $$2$$, then $$n - 3$$ choices for where to send $$3$$...)
Another way to count is more complicated: for each $$1 \le j \le n - 2$$, choose $$j - 1$$ elements of $$\{2, \dotsc, n - 1\}$$ to keep fixed (note a permutation cannot fix all but one element), and then choose a derangement of the remaining $$n - j$$ elements of $$\{1, \dotsc, n - 1\}$$. This counts each of the permutations we're interested in exactly once, so it follows that $$\begin{equation*} \sum_{j = 1}^{n - 2} c_{n - j} \binom{n - 2}{j - 1} = (n - 2)!(n - 2). \end{equation*}$$