For every pair $(p, k)$ of nonzero natural numbers, we denote by $S(p, k)$ the number of surjections from $\llbracket 1, p \rrbracket$ to $\llbracket 1, k \rrbracket$. Consistently, for every $p \in \mathbb{N}^*$, we set $S(p, 0) = 0$.
Simplify as much as possible the following expressions:
$$\sum_{k=0}^{n} (-1)^{n-k} \binom{n}{k} k^n \quad \text{and} \quad \sum_{k=0}^{n} (-1)^{n-k} \binom{n}{k} k^{n+1}$$