Prove $\sum_{k = 0}^{n}(-1)^{n - k} \binom{n}{k} \cdot k^n = n!$ and $\sum_{k = 0}^{n}(-1)^{n - k} \binom{n}{k} \cdot k^m = 0$

Consider the exponential generating function

$$ f(x) = \sum_{m=0}^{\infty} \frac{A_m}{m!} x^m, \qquad A_m := \sum_{k=0}^{n} (-1)^{n-k}\binom{n}{k} k^m. $$

Then it follows that

\begin{align*} f(x) &= \sum_{k=0}^{n} (-1)^{n-k}\binom{n}{k} \sum_{m=0}^{\infty} \frac{(kx)^m}{m!} \\ &= \sum_{k=0}^{n} (-1)^{n-k}\binom{n}{k} e^{kx} \\ &= (e^x - 1)^n. \end{align*}

Writing $e^x - 1 = x g(x)$ for $g(x) = \sum_{j=0}^{\infty} \frac{x^j}{(j+1)!}$,

$$ f(x) = x^n g(x)^n . $$

In particular, the Maclaurin series for $f$ about $x = 0$ begins with $ f(x) = x^n + \cdots $, which translates to the desired fact:

$$ A_m = \begin{cases} 0, &\text{if $m < n$}, \\ n!, &\text{if $m = n$}. \end{cases} $$

In general, this computation shows that

$$ A_m = \sum_{\substack{j_1\geq1,\dots,j_n\geq1 \\ j_1+\cdots+j_n=m}} \binom{m}{j_1,\dots,j_n} \tag{1} $$

for all $m = 0, 1, 2, \dots $.

Alternatively, here is an extended version of @Trevor Gunn's answer: Let $[n] = \{1, \dots, n\}$. Then define

$$ X_i = \{ f \in [n]^{[m]} : i \notin \operatorname{range}(f) \} = ([n]\setminus\{i\})^{[m]} $$

for each $i = 1, 2, \dots, n$ and

$$ X_{\varnothing} = [n]^{[m]}, \qquad X_I = \bigcap_{i\in I} X_i, $$

for each non-empty $I \subseteq [n]$. Then $A_m$ can be written as

$$ A_m = \sum_{k=0}^{n} (-1)^k \binom{n}{k} (n-k)^m = \sum_{I\subseteq[n]} (-1)^{\left| I \right|} \left| X_I \right|. $$

Now by the inclusion-exclusion principle, it follows that

\begin{align*} A_m &= \left| \{ f \in [n]^{[m]} : f \notin X_1 \cup \cdots \cup X_n \} \right| \\ &= \text{[# of surjections from $[m]$ to $[n]$]}. \tag{2} \end{align*}

Note that this formula also matches the formula $\text{(1)}$ in the first solution. Now the desired answer easily follows from $\text{(2)}$.

Here are some hints:

1. The $(-1)^{n - k}$ suggests to interpret this as inclusion exclusion.
2. $k^n$ is the number of functions from an $n$ element set to a $k$ element set
3. $n!$ is the number of functions from an $n$ element set to an $n$ element set which are injections/surjections/bijections

So you can think of the term $\binom{n}{k}k^n$ as the number of functions from $\{1,\dots,n\} \to \{1,\dots,n\}$ whose image is contained in a chosen $k$-element subset of the codomain.

This problem has appeared a number of times. Using coefficient extractors we have

$$\sum_{k=0}^n (-1)^{n-k} {n\choose k} k^m \\ = m! [z^m] \sum_{k=0}^n (-1)^{n-k} {n\choose k} \exp(kz) \\ = m! [z^m] (\exp(z)-1)^n.$$

Now $\exp(z)-1 = z+\cdots$ so $(\exp(z)-1)^n = z^n + \cdots$ and therefore the coefficient extractor returns zero when $m\lt n$ and one when $m=n$, which is the claim.