$\sum_{k=0}^n\binom{n}{k}(-1)^k(k+1)$ Am I true? $\displaystyle\sum_{k=0}^n\binom{n}{k}(-1)^k(k+1)=\underbrace{\displaystyle\sum_{k=0}^n\binom{n}{k}(-1)^k}_{(1+(-1))^n=0}+\displaystyle\sum_{k=0}^n\binom{n}{k}(-1)^k\;k$
And we have , $\displaystyle\sum_{k=0}^n\binom{n}{k}(-1)^k\;k$
I thought that I know $\dbinom{n}{m}=\dfrac{n}{m}\dbinom{n-1}{m-1}$
Thus, I can say, $\displaystyle\sum_{k=0}^n\binom{n}{k}(-1)^k\;k=\displaystyle\sum_{k=0}^n\binom{n-1}{k-1}(-1)^k\;n=\underbrace{\displaystyle\sum_{k=1}^n\binom{n-1}{k-1}(-1)^k}_{(1+(-1))^{n-1}=0}\;n+\underbrace{\dbinom{n-1}{-1}}_{\text{A}}n$
Lets evaluate $A$;
$A=\dbinom{n-1}{-1}=\dfrac{(n-1)!}{(-1)!n!}$
We know what $(-1)!$ equals to $\infty$  "Source:https://mathoverflow.net/questions/10124"
Then I think I can say this $A=0$ and  $\quad \displaystyle\sum_{k=0}^n\binom{n}{k}(-1)^k\;k=\displaystyle\sum_{k=1}^n\binom{n-1}{k-1}(-1)^k\;n+\dbinom{n-1}{-1}n=0$
Then Can I say $\displaystyle\sum_{k=0}^n\binom{n}{k}(-1)^k(k+1)$ also equals to zero? Am I wrong?
 A: Your derivation is probably the most straightforward computational derivation. A combinatorial argument is also possible.
Let $S=\{0,1,\ldots,n\}$. For $k\in[n]$ let $\mathscr{A}_k$ be the family of one-element subsets of $S$ not containing $k$. If $\varnothing\ne I\subseteq[n]$, then
$$\left|\,\bigcap_{k\in I}\mathscr{A}_k\,\right|=n+1-|I|\;,$$
so by the inclusion-exclusion principle
$$\begin{align*}
\left|\,\bigcup_{k\in[n]}\mathscr{A}_k\,\right|&=\sum_{\varnothing\ne I\subseteq[n]}(-1)^{|I|+1}(n+1-|I|)\\
&=\sum_{k=1}^n(-1)^{k+1}\binom{n}k(n+1-k)\;.
\end{align*}$$
This is the number of one-element subsets of $S$ that exclude at least one member of $[n]$. Provided that $n\ge 2$, every one-element subset of $S$ excludes at least one member of $[n]$, and there are $n+1$ one-element subsets of $S$, so for $n\ge 2$ we have
$$\begin{align*}
0&=n+1-\sum_{k=1}^n(-1)^{k+1}\binom{n}k(n+1-k)\\
&=n+1+\sum_{k=1}^n(-1)^k\binom{n}k(n+1-k)\\
&=\sum_{k=0}^n(-1)^k\binom{n}k(n+1-k)\\
&=\sum_{k=0}^n(-1)^{n-k}\binom{n}k(k+1)\\
&=(-1)^n\sum_{k=0}^n(-1)^k\binom{n}k(k+1)\;,
\end{align*}$$
and hence
$$\sum_{k=0}^n(-1)^k\binom{n}k(k+1)=0\;.$$
Note that the result fails for $n=1$: the sum is $-1$ when $n=1$. This is because when when $n=1$, only one of the two one-element subsets of $S=\{0,1\}$ excludes at least one member of $[1]=\{1\}$. (It also fails for $n=0$, because there are no elements of $[0]=\varnothing$ to be excluded, and the single one-element subset of $\{0\}$ therefore does not exclude any of them.)
A: $$
I = \sum_{k=0}^{n}\left(\matrix{n\\k}\right)a^{k+1} \implies \frac{dI}{da} = \sum_{k=0}^{n}\left(\matrix{n\\k}\right)a^{k}(k+1)
$$
so solving the l.h.s allows for solving your series.
$$
a\sum_{k=0}^{n}\left(\matrix{n\\k}\right)a^{k} =a(1+a)^n = I
$$x
so 
$$
\frac{dI}{da} = (1+a)^n +an(1+a)^{n-1}
$$
using $a=-1$ we find
$$
\frac{dI}{da} = 0 = \sum_{k=0}^{n}\left(\matrix{n\\k}\right)(-1)^{k}(k+1)
$$
So you are correct with answer.
Now to your point revolving $(-1)! =\infty$ I am having more trouble with since $(-1)!$ is not really defined in the traditional sense.
