Alternate sum of Worpitzky numbers (similar to Stirling numbers) Considering the Worpitzky numbers $W(n, k)$, defined in terms of the Stirling numbers of the second kind $S(n, k)$ as
$$
W(n, k) = k!S(n, k) = \sum_{i=0}^k (-)^i \binom{k}{i} (k-i)^n
$$
Begin Edit
It is the wrong definition for the Worpitzky numbers (see comments), sorry for that (it was late). The real definition is 
$$
W(n, k) = k!S(n+1, k+1)
$$
I do not change the rest of the question to not be confusing with respect to the answers.
End Edit
I suspect the alternate sum of $W(n, k)$ for fixed $n$ (number of elements in the set) is given by
$$
\sum_{k=1}^n(-)^kW(n, k) = (-)^n
$$
This question is a follow up to my previous question on the subject. In that context, I could verify
$$
\sum_{k=1}^1(-)^kW(1, k) = -1
$$
$$
\sum_{k=1}^2 (-)^k W(2, k) = -1 + 2 = 1
$$
$$
\sum_{k=1}^3 (-)^k W(3, k) = -1 + 6 - 6 = -1
$$
$$
\sum_{k=1}^4 (-)^k W(4, k) = -1 + 14 - 36 + 24 = 1
$$
But I can't find anything in my brain or googlable literature to generalize this.
 A: Using the definition by Wikipedia 
I and Wikipedia 
II we have
$$W_{n,k} = k! {n+1\brace k+1}.$$
This gives for the sum
$$\sum_{k=1}^n (-1)^k W_{n,k} =
\sum_{k=1}^n (-1)^k k! {n+1\brace k+1} =
-1 + \sum_{k=0}^n (-1)^k k! {n+1\brace k+1}
\\ = -1 + \sum_{k=0}^n (-1)^k k! (n+1)! [z^{n+1}]
\frac{(\exp(z)-1)^{k+1}}{(k+1)!}
\\ = -1 + n! [z^{n}]
\sum_{k=0}^n (-1)^k k!
\frac{(\exp(z)-1)^k}{k!} \exp(z)
\\ = -1 + n! [z^{n}] \exp(z)
\sum_{k=0}^n (-1)^k (\exp(z)-1)^k
\\ = -1 + n! [z^{n}] \exp(z)
\sum_{k=0}^n (1-\exp(z))^k
\\ = -1 + n! [z^{n}] \exp(z)
\frac{1-(1-\exp(z))^{n+1}}{1-(1-\exp(z))}
\\ = -1 + n! [z^{n}]
(1-(1-\exp(z))^{n+1}).$$
Now since $1-\exp(z) = -z - \cdots$ we have $[z^n] (1-\exp(z))^{n+1} =
0$ so that
$$\sum_{k=1}^n (-1)^k W_{n,k} = -1.$$
If on the other hand we are working with the sum
$$\sum_{k=1}^n (-1)^k k!  {n\brace k}$$
we obtain
$$n! [z^n] \sum_{k=1}^n (-1)^k (\exp(z)-1)^k$$
which is for $n\ge 1$
$$n! [z^n] \sum_{k=0}^n (-1)^k (\exp(z)-1)^k
= n! [z^n] \sum_{k=0}^n (1-\exp(z))^k
\\ = n! [z^n] \exp(-z) (1-(1-\exp(z))^{n+1})
= n! [z^n] \exp(-z) = (-1)^n.$$
A: Here's an alternative to Marko's answer that doesn't use exponential generating functions.  As implied in his answer and my comment, your question is not about the Worpitzky numbers $k! S(n+1,k+1)$ (using your notation for the Stirling numbers of the second kind) rather the numbers $k! S(n,k)$ which are A019538.  As a triangle $T(n,k)$ for $1 \le k\le n$, they begin
\begin{array}{ccccc}
1 \\ 1 & 2 \\ 1 & 6 & 6 \\ 1 & 14 & 36 & 24 \\ 1 & 30 & 150 & 240 & 120
\end{array}
Note the relation row to row is similar to Pascal's triangle but weighted by column: $$T(n,k) = k\left(T(n-1,k-1)+T(n-1,k)\right).$$  That's enough to prove your alternating sum identity using induction.  For instance, for the fourth row,
\begin{align}
-T(4,1)&+T(4,2)-T(4,3)+T(4,4) \\
& = -1 + 2(T(3,1)+T(3,2)) - 3(T(3,2)+T(3,3))+4(T(3,3)+T(3,4)) \\
& = -1 + T(3,1) + [T(3,1)-T(3,2)+T(3,3)] \\
& = -1 + 1 - [-T(3,1)+T(3,2)-T(3,3)] \\
& = 0 -(-1) \\
& = 1
\end{align}
where the bracketed expression in the third to last line is the alternating third row sum.
A: Alternative approach: Denote $[n]=\{1,2,\ldots, n\}$. We have for all $x\in\mathbb{C}$,
$$
x^n=\sum_{k=0}^n S(n,k) (x)_k
$$
where $(x)_k=x(x-1)\cdots (x-k+1)$ is the falling factorial.
Proof. It suffices to prove the identity for all natural number $x$. The LHS is the number of $x$-colorings of $[n]$. The summand in the RHS,
$$
S(n,k) k!\binom xk, 
$$
is the number of ways to select $k$ colors from $x$ colors and color $[n]$ using all selected colors.
Proof of the alternating sum identity:
Put $x=-1$ into the identity, we have
$$
(-1)^n=\sum_{k=0}^nS(n,k)k! (-1)^k.  
$$
For Worpitzky numbers $W(n,k)=S(n+1,k+1)k!$, we modify the identity by $k\binom xk = x \binom{x-1}{k-1}$. We have
$$
x^n=\sum_{k=1}^nS(n,k)(k-1)!x\binom{x-1}{k-1}=\sum_{k=0}^{n-1}S(n,k+1)k!x\binom{x-1}k.
$$
Replacing $n$ by $n+1$, we have
$$
x^{n+1}=\sum_{k=0}^nS(n+1,k+1)k!x\binom{x-1}k.
$$
We drop $x$ and replace $x$ by $x+1$, we have
$$
(x+1)^n=\sum_{k=0}^nS(n+1,k+1)k!\binom xk.
$$
Put $x=-1$ above, then
$$
0=\sum_{k=0}^nS(n+1,k+1)k!(-1)^k.
$$
Isolating $k=0$ in the summation, we obtain the alternating sum identity for Worpitzky numbers.
