sum of alternating binomial 
Compute the sum $\sum_{k=0}^{n}(-1)^k k^n\binom{n}{k} $

I've seen a solution along the following lines here, page 3: 
Consider $(1+x)^n=\sum_{k=0}^{n}\binom{n}{k}x^k$. ($\star$) We prove by induction that $\sum_{k=0}^{n}(-1)^k k^t\binom{n}{k}=0$ for $t< n$. We prove this by differentiating ($\star$) t times, setting $x=-1$ and using the inductive step.
Now if we differentiate $(\star)$ n times we get: $n!=\sum_{k=0}^{n} k \cdot (k-1) \dots \cdot (k-(n-1)) \binom{n}{k}x^{k-n}=\sum_{k=0}^{n} k^n\binom{n}{k}x^{k-n}$ (by the inductive step).
So $n!=\sum_{k=0}^{n} k^n\binom{n}{k}x^{k-n}$ and by setting $x=-1$ and multiplying by $(-1)^n$ we get $\sum_{k=0}^{n}(-1)^k k^n\binom{n}{k}=n! (-1)^n $.
My question is, starting from $(1+x)^n=\sum_{k=0}^{n}\binom{n}{k}x^k$, if we differentiate it n times, most of the RHS terms will vanish ,leaving us with $n!=n!$, not $n!=\sum_{k=0}^{n} k \cdot (k-1) \dots \cdot (k-(n-1)) \binom{n}{k}x^{k-n}$. How is that a valid step and also how can the sum in question be evaluated?
 A: Here is an answer based upon generating functions which might provide an alternate solution. It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write for instance
\begin{align*}
n![x^n]e^{kx}=k^n\tag{1}
\end{align*}

We obtain
  \begin{align*}
\color{blue}{\sum_{k=0}^n\binom{n}{k}(-1)^kk^n}
&=\sum_{k=0}^n\binom{n}{k}(-1)^k n![x^n]e^{kx}\tag{2}\\
&=n![x^n]\sum_{k=0}^n\binom{n}{k}\left(-e^x\right)^k\tag{3}\\
&=n![x^n](1-e^x)^n\tag{4}\\
&=n![x^n]\left(-x-\frac{x^2}{2!}-\frac{x^3}{3!}-\cdots\right)^n\\
&\,\,\color{blue}{=(-1)^nn!}\tag{5}
\end{align*}

Comment:


*

*In (2) we apply the coefficient of operator according to (1).

*In (3) we use the linearity of the coefficient of operator.

*In (4) we apply the binomial theorem.

*In (5) we select the coefficient of $x^n$.
A: Define the linear functional transform $\;D_x[f(x)] := x f'(x).\;$ Clearly,
 $\;D_x[x^n] = nx^n,\;$ also
 $\;D_x[(x+y)^n] = nx(x+y)^{n-1}.\;$
The binomial theorem states that
 $\;(x+y)^n = \sum_{k=0}^n {n \choose k}x^k y^{n-k}.\;$
Apply $\;D_x\;$ to both sides of the equation gives
 $\;nx(x+y)^{n-1} = \sum_{k=0}^n {n \choose k}\;k\; x^k y^{n-k}.\;$
Apply $\;D_x\;$ to both sides several times gives a polynomial in $\;x,y\;$ whose coefficients (up to alternating sign) are OEIS triangular sequence A258773. The right side is
 $\; \sum_{k=0}^n {n \choose k}\;k^n\; x^k y^{n-k}.\;$
Let $\;x=-1,\; y=1\;$ to get
 $\;n!(-1)^n = \sum_{k=0}^n {n \choose k}\;k^n\;(-1)^k,$
where the left side comes from the OEIS sequence entry.
I also used  $\;D_x\;$ in my answer to a MSE question 2772848 on property of Bernstein polynomial.
A: First some preliminary identities:
We use the identity
$$
k^n=\sum_{m=0}^n{n\brace m}(k)_m\tag{1}
$$
where the braces indicate stirling numbers of the second kind and $(k)_m=k(k-1)\dotsb (k-m+1)$ indicates the falling factorial of length $m$. Also note that
$$
(1-x)^n=\sum_{k=0}^n\binom{n}{k} (-1)^kx^k
$$
implies that
$$
\sum_{k=0}^n\binom{n}{k} (k)_m(-1)^k=\frac{d^m}{dx^m}(1-x)^n\bigg|_{x=1}=(n)_m(-1)^m\delta_{n,m}\quad (0\leq m\leq n)\tag{2}
$$
where $\delta$ indicates the Kronecker Delta. Now we can proceed with the problem. Indeed,
$$
\begin{align}
\sum_{k=0}^n\binom{n}{k}(-1)^kk^n
&\stackrel{(1)}{=}\sum_{k=0}^n\binom{n}{k}(-1)^k\left(
\sum_{m=0}^n{n\brace m}(k)_m
\right)\\
&=\sum_{m=0}^n{n\brace m}\left(\sum_{k=0}^n(-1)^k\binom{n}{k}(k)_m\right)\\
&\stackrel{(2)}{=}\sum_{m=0}^n{n\brace m}(n)_m(-1)^m\delta_{n,m}\\
&=(-1)^nn!
\end{align}
$$
as desired.
