Taking the derivative $n$ times and setting $x = 0$ gives you the coefficient of $x^n$ in the exponential generating function:
$$ f(x) = \sum_{k = 0}^\infty f^{(k)}(0) \frac{x^k}{k!}. $$
It is typical to write this as in terms of the coefficient operator $\left[ \frac{x^n}{n!} \right]$ which when applied to a power series $f(x)$ gives you the coefficient of $\frac{x^n}{n!}$ in $f(x)$. In other words
$$\left[ \frac{x^n}{n!} \right] \sum_{k = 0}^\infty a_k \frac{x^k}{k!} = a_n. $$
Then
\begin{align}
\sum_{k = 0}^n \binom{n}{k} (-1)^{n - k}k^n &= \sum_{k = 0}^n \binom{n}{k} (-1)^{n - k} \left[ \frac{x^n}{n!} \right] e^{kx} \\
&= \left[ \frac{x^n}{n!} \right]\sum_{k = 0}^n \binom{n}{k} (-1)^{n - k} e^{kx} \\
&= \left[ \frac{x^n}{n!} \right] (e^x - 1)^n.
\end{align}
In the theory of combinatorial species, $e^x - 1$ is the (exponential) generating function for non-empty sets. That is, of the map which takes in a set $X$ and spits out the set
$$ \mathcal{E}(X) = \begin{cases} \{X\} & \text{if } X \ne \emptyset \\ \emptyset & \text{if } X = \emptyset \end{cases}. $$
Notice that
$$ |\mathcal{E}(X)| = \begin{cases} 1 & \text{if } X \ne \emptyset \\ 0 & \text{if } X = \emptyset \end{cases} $$
We define the generating function for $\mathcal{E}$ to be
$$ \sum_{n = 0}^\infty |\mathcal{E}([n])| \frac{x^n}{n!} = e^x - 1 $$
where $[n] = \{1, 2, \dots, n\}$.
If $\mathcal{F}$ is a combinatorial species then we define the power $\mathcal{F}^k$ to be the map
$$ \mathcal{F}^k(X) = \bigsqcup_{(S_1,\dots,S_k)} \mathcal{F}(S_1) \times \dots \times \mathcal{F}(S_k) \tag{1}$$
where the union is over all (ordered) partitions of $X$ into $k$ subsets $S_1,\dots,S_k$. That is, $S_1 \cup \dots \cup S_k = X$ and $S_1,\dots,S_k$ are disjoint.
One can show that if $F(x) = \sum_n |\mathcal F([n])| x^n/n!$ is the generating function for $\mathcal{F}$ then $F(x)^k$ is the generating function for $\mathcal{F}^k$.
Bringing this back to $\mathcal{E}^n$ we can say that
$$ \left[ \frac{x^n}{n!} \right] (e^x - 1)^n = |\mathcal{E}^n([n])| $$
is the number of ways of partitioning $[n]$ into $n$-non-empty sets because only the terms in $(1)$ in which $S_1,\dots,S_n$ are all non-empty will contribute to the union.
Partitioning $[n]$ into $n$ singletons is the same thing as a permutation of $n$ and there are $n!$ of these. Hence
$$n! = \left[ \frac{x^n}{n!} \right] (e^x - 1)^n = \sum_{k = 0}^n \binom{n}{k} (-1)^{n - k}k^n. $$