Perhaps you are unaware about this approach, but we can solve this problem using generating functions and the fundamental recurrence for the Stirling numbers of the second kind. In fact, this is done as an example in the fantastic book by Herbert Wilf, Generatingfunctionology.
Let, $S(n, k) = {n \brace k}$ be the Stirling number of the second kind.
We have, for $n, k \gt 0$ that, $$S(n, k) = S(n-1, k-1) + k \cdot S(n-1, k)$$ with the initial value $S(0,0) = 1$.
Define the generating function $F_k(x) = \displaystyle \sum_{n \ge 0} S(n, k) x^n$
Then, from the recurrence relation, we get after summing over $n$,
$$F_k(x) = xF_{k-1}(x) + kxF_k(x)$$
from which we have,
$$F_k(x) = \dfrac{x}{1 - kx}F_{k-1}(x) = \displaystyle \prod_{r = 1}^k \dfrac{x}{1 - rx}$$
since the initial value is $F_0(x) = 1$.
Now, we need to do a partial fraction expansion of the denominator in the product to get the explicit formula.
Let,
$\displaystyle \prod_{r = 1}^k \dfrac{1}{1 - rx} = \displaystyle \sum_{j = 1}^k \dfrac{p_j}{1 - jx}$
Cross-multiply one factor at a time and set that factor to zero. We easily get that, $$p_j = \displaystyle \prod_{r = 1}^{k; r \ne j} \dfrac{1}{1 - \frac{r}{j}} = (-1)^{k - j}\dfrac{j^k}{j! \cdot (k-j)!}$$
Hence, we have,
$$F_k(x) = \displaystyle \sum_{n \ge 0} S(n, k) x^n = x^k \displaystyle \prod_{r = 1}^k \dfrac{1}{1 - rx}$$
$$\implies F_k(x) = x^k \displaystyle \sum_{j = 1}^k \dfrac{p_j}{1 - jx}$$
$$\implies F_k(x) = x^k \displaystyle \sum_{j = 1}^k p_j(1 + jx + j^2 x^2 + \dots)$$
$$\implies F_k(x) = \displaystyle \sum_{j = 1}^k p_j(x^k + j x^{k+1} + \dots)$$
The coefficient of $x^{k + r}$ on the right hand side is nothing but $\displaystyle \sum_{j = 1}^k p_j \cdot j^r$. Setting $k + r = n$ and using the definition of the generating function, we have,
$$S(n, k) = \displaystyle \sum_{j = 1}^k p_j \cdot j^{n - k}$$
And using the expression for $p_j$, we finally get,
$$S(n, k) = \displaystyle \sum_{j = 1}^k (-1)^{k - j}\dfrac{j^n}{j! \cdot (k-j)!}$$
Though it can't be said that computing the Stirling numbers using this formula is very efficient, the advantage is that this formula puts all problems like the one in Brualdi to rest at once. Hope it helps.
Best Answer
If the permutation $\pi$ of $[n]$ is a derangement, $\langle\pi(1),\pi(2),\ldots,\pi(n)\rangle$ is an SDR: for each $k\in[n]$ we have $\pi(k)\ne k$, so $\pi(k)\in A_k$. The converse is really just the same idea: if $\langle a_1,a_2,\ldots,a_n\rangle$ is an SDR, then the permutation $\pi:[n]\to[n]:k\mapsto a_k$ is a derangement, since $a_k\notin A_k$ for $k\in[n]$.