Exponential Generating Function For Derangements
Here’s one way. Start with the recurrence $d_{n+1}=nd_n+nd_{n-1}$, where $d_n$ is the number of derangements of $[n]=\{1,\dots,n\}$. Multiply by $\frac{x^n}{n!}$ and sum over $n\ge 0$:
$$\sum_{n\ge 0}d_{n+1}\frac{x^n}{n!}=\sum_{n\ge 0}nd_n\frac{x^n}{n!}+\sum_{n\ge 0}nd_{n-1}\frac{x^n}{n!}\;.\tag{1}$$
(Here I make the assumption that $d_{-1}=0$: this is consistent with the recurrence, so it causes no problems.) Let $$D(x)=\sum_{n\ge 0}d_n\frac{x^n}{n!}$$ be the exponential generating function in question. Then $$D\,'(x)=\sum_{n\ge 0}nd_n\frac{x^{n-1}}{n!}=\sum_{n\ge 1}d_n\frac{x^{n-1}}{(n-1)!}=\sum_{n\ge 0}d_{n+1}\frac{x^n}{n!}\;,\tag{2}$$
$$xD\,'(x)=x\sum_{n\ge 0}nd_n\frac{x^{n-1}}{n!}=\sum_{n\ge 0}nd_n\frac{x^n}{n!}\;,\tag{3}$$
and $$xD(x)=\sum_{n\ge 0}d_n\frac{x^{n+1}}{n!}=\sum_{n\ge 0}(n+1)d_n\frac{x^{n+1}}{(n+1)!}=\sum_{n\ge 1}nd_{n-1}\frac{x^n}{n!}=\sum_{n\ge 0}nd_{n-1}\frac{x^n}{n!}\;,\tag{4}$$ since by convention $d_{-1}=0$.
Compare $(2),(3)$, and $(4)$ with $(1)$, and you’ll see that
$$D\,'(x)=xD\,'(x)+xD(x)\;,$$ or $$(1-x)D\,'(x)=xD(x)\;.$$ This is a separable differential equation,
$$\frac{D\,'(x)}{D(x)}=\frac{x}{1-x}=-1+\frac1{1-x}\;,$$
which you can now solve for $D(x)$ by first-year calculus.
Here’s another way, quicker but sneakier. For any particular set $K$ of $k$ elements of $[n]$ there are $d_{n-k}$ permutations of $[n]$ that have $K$ as their set of fixed points. There are $\binom{n}k$ such subsets $K$, so there are $\binom{n}kd_{n-k}$ permutations of $[n]$ with exactly $k$ fixed points. Since there are $n!$ permutations of $[n]$ altogether, $$\sum_{k=0}^n\binom{n}kd_{n-k}=n!\;.\tag{5}$$ The lefthand side of $(5)$ is the $n$-th term of the binomial convolution of the sequences $\langle 1,1,1,\dots\rangle$ and $\langle d_n:n\in\Bbb N\rangle$, so the exponential generating function (egf) of the sequence
$$\left\langle\sum_{k=0}^n\binom{n}kd_{n-k}:n\in\Bbb N\right\rangle=\langle n!:n\in\Bbb N\rangle$$
is the product of the egfs of $\langle 1,1,1,\dots\rangle$ and $\langle d_n:n\in\Bbb N\rangle$. Clearly $$\sum_{n\ge 0}n!\frac{x^n}{n!}=\sum_{n\ge 0}x^n=\frac1{1-x}$$ and $$\sum_{n\ge 0}1\cdot\frac{x^n}{n!}=e^x\;,$$ so $$e^xD(x)=\frac1{1-x}\;,$$ and $$D(x)=\frac{e^{-x}}{1-x}\;.$$
There's a shorter way even than the two in Brian's nice answer. Starting from the recurrence $D_n = n D_{n-1} + (-1)^n$, multiply by $x^n/n!$ and sum over $n \geq 1$ to get \begin{align} &\sum_{n \geq 1} D_n \frac{x^n}{n!} = \sum_{n \geq 1} n D_{n-1} \frac{x^n}{n!} + \sum_{n \geq 1} (-1)^n \frac{x^n}{n!} \\ \implies & G_D(x) - 1 = x \sum_{n \geq 1} D_{n-1} \frac{x^{n-1}}{(n-1)!} + e^{-x} - 1, \text{ as $D_0 = 1$} \\ \implies & G_D(x) = x G_D(x) + e^{-x}, \end{align} which tells you that the exponential generating function is $$G_D(x) = \frac{e^{-x}}{1-x}.$$
Another way: There are $n!$ permutations in all. A permutation with $k$ fixed points shuffles the other $n - k$ elements around without fixed points, the $k$ fixed points can be selected in $\binom{n}{k}$ ways. So: $$ n! = \sum_{0 \le k \le n} \binom{n}{k} d_{n -k} $$ This is a binomial convolution, using Mike Spivey's notation that gives: $$ \begin{align*} \frac{1}{1 - x} &= G_D(x) e^x \\ G_D(x) &= \frac{e^{-x}}{1 - x} \end{align*} $$