I have a problem understanding the proof of Rencontres numbers (Derangements) I understand the whole concept of Rencontres numbers but I can't understand how to prove this equation
$$D_{n,0}=\left[\frac{n!}{e}\right]$$
where $[\cdot]$ denotes the rounding function (i.e., $[x]$ is the integer nearest to $x$). This equation that I wrote comes from solving the following recursion, but I don't understand how exactly the author calculated this recursion.
$$\begin {align*} 
D_{n+2,0} & =(n+1)(D_{n+1,0}+D_{n,0}) \\
D_{0,0} & = 1 \\
D_{1,0} & = 0 
\end {align*}
$$

(This argument is adapted from page 195 of Concrete Mathematics, Second Edition)
We start with the more conventional representation for the Rencontres number (subfactorial):
$$D_{n,0}=!n=n!\sum_{k=0}^n \frac{(-1)^k}{k!}$$
We also know that
$$\frac{n!}{e}=n!\sum_{k=0}^\infty \frac{(-1)^k}{k!}$$
The difference is
$$\begin{align*}\frac{n!}{e}-!n&=n!\sum_{k=n+1}^\infty \frac{(-1)^k}{k!}\\&=\frac{(-1)^{n+1}}{n+1}\left(1-\frac1{n+2}+\frac1{(n+2)(n+3)}-\cdots\right)\end{align*}$$
and since
$$\frac1{n+2} \leq \left|\frac{n!}{e}-!n\right| \leq \frac1{n+1}$$
along with knowing that $!n$ is an integer, rounding $n!/e$ to the nearest integer gives the subfactorial.

We have
$\small \begin{align}(n+2)!\sum_{k=0}^{n+2} \frac{(-1)^k}{k!}&=(n+1)\left[(n+1)!\sum_{k=0}^{n+1} \frac{(-1)^k}{k!}+n!\sum_{k=0}^n \frac{(-1)^k}{k!}\right]\\(n+2)(n+1)\sum_{k=0}^{n+2} \frac{(-1)^k}{k!}&=(n+1)\left[(n+1)\sum_{k=0}^{n+1} \frac{(-1)^k}{k!}+\sum_{k=0}^n \frac{(-1)^k}{k!}\right]\\(n+2)\sum_{k=0}^{n+2} \frac{(-1)^k}{k!}&=(n+1)\sum_{k=0}^{n+1} \frac{(-1)^k}{k!}+\sum_{k=0}^n \frac{(-1)^k}{k!}\\(n+2)\left(\frac{(-1)^{n+2}}{(n+2)!}+\frac{(-1)^{n+1}}{(n+1)!}+\sum_{k=0}^n \frac{(-1)^k}{k!}\right)&=\frac{(-1)^{n+1}}{n!}+(n+1)\sum_{k=0}^n \frac{(-1)^k}{k!}+\sum_{k=0}^n \frac{(-1)^k}{k!}\\(n+2)\left(\frac{(-1)^n}{(n+2)!}+\frac{(-1)^{n+1}}{(n+1)!}\right)+(n+2)\sum_{k=0}^n \frac{(-1)^k}{k!}&=\frac{(-1)^{n+1}}{n!}+(n+1)\sum_{k=0}^n \frac{(-1)^k}{k!}+\sum_{k=0}^n \frac{(-1)^k}{k!}\\(n+2)\left(\frac{(-1)^n}{(n+2)!}+\frac{(-1)^{n+1}}{(n+1)!}\right)&=\frac{(-1)^{n+1}}{n!}\\(-1)^n+(-1)^{n+1}(n+2)&=(-1)^{n+1}(n+1)\\1-(n+2)&=-(n+1)\end{align}$
and the last bit is easily established, thus proving the recursion relation for the Rencontres numbers.
