An infinite set of identities using Stirling numbers 1st kind - are they all zero? I have the following set of series involving the Stirling numbers 1'st kind and binomials, which can be understood as a set of dot-products of row- and column-vectors of two infinite matrices (where R and C indicate rows and columns, beginning at zero):
$$ w_{R,C} =\sum_{k=\max(R,C)}^\infty (-1)^k {s_1(1+k,1+k-R)\over k!} \cdot (-1)^C (1+C)^k \cdot \binom {1+k}{1+C} $$
I've tested this heuristically for several R and C and always approximated zero; also wolfram-alpha can evaluate this explicitely to zero if feeded with      

sum (-1)^k * StirlingS1(k+1,1+k-R)/k! * (1+C)^k * binomial(1+k,1+C), for k=max(C,R) to infty      

where we replace $C$, $R$ and $\max(C,R)$ with actual values.     
However, I've no option to let wolfram-alpha answer this in general.      
I've proved this for $C=0,1,2$ and the first few $R$ using exponential generating functions, but again, a general proof is out of reach for me (possibly I'm overlooking something trivial like telescoping...), so I ask for help here.     

The convention for Stirling numbers first kind as in Math'ica, indexes beginning at zero:       

$ \small \qquad \qquad \begin{array} {rrrrr}
 1 & . & . & . & . & . \\\
 0 & 1 & . & . & . & . \\\
 0 & -1 & 1 & . & . & . \\\
 0 & 2 & -3 & 1 & . & . \\\
 0 & -6 & 11 & -6 & 1 & . \\\
 0 & 24 & -50 & 35 & -10 & 1
 \end{array} $

If some background is of interest: here are the questions on MSE
https://math.stackexchange.com/questions/16228    // question of some user which motivated me to look at an example
https://math.stackexchange.com/questions/89853    // my follow-up question dealing with the current problem
and a more worked out treatize on this in a pdf-file 
 http://go.helms-net.de/math/divers/InverseNullmatrix.pdf

[update] Hmm, after 1 1/2 years I've looked at the question again and still do not have an idea how to construct a proof for the whole set of identities. To possibly stimulate helpful answers here I'll insert pictures of the matrices - perhaps it helps to get an immediate idea when the patterns are more visible/obvious than in the bare formula above. 
This is (the top-left-segment of) the matrix $M$ in question.           
 
This are the L and D factors of the L D U-decomposition. Because it seems convenient to recognize familiar numbers I've documented the product LD = L D 
 
This is the U factor:                    
 

This is the reciprocal of U (call it UI):             
 
This is the reciprocal of LD (call it LDI):             
 
and in the limit for infinite size of the UI and LDI, the product UI * LDI = MI = 0  by hypothese. 
 
Here are the matrices UI and LDI in a near-symbolic display, the coefficients $s1[r,c]$ are the Stirling numbers first kind.
 

 
Reformulating the dotproducts using their exponential generating functions it is not difficult to prove the identities for a couple of examples.
But what is missing is the proof for the full set of dotproducts.
[/update]
 A: Cancelling a few terms, you see that we want to show that
$$
\sum_{k=C}^\infty (-1-C)^k {s_1(1+k,1+k-R)(1+k)\over (k-C)!}=0.
$$
Let us further simplify this equation via the coordinate transformations $C\leftarrow -C-1$ and $k\leftarrow k+1$. We then want to prove that
$$
I_{R,C}:=\sum_{k=-C}^\infty \frac{C^k}{(k+C)!}\cdot ks_1(k,k-R)=0
$$
for all $C\leq-1$ and $R\geq0$.
Recall the recurrence relation
$$ks_1(k,k-R)=s_1(k,k-R-1)-s_1(k+1,k-R).$$
Using summation by parts, we obtain
$$
\begin{aligned}
I_{R,C}
&=\sum_{k=-C}^\infty \frac{C^k}{(k+C)!}\cdot (s_1(k,k-R-1)-s_1(k+1,k-R))\\
&=B+\sum_{k=-C}^\infty \left(\frac{C^{k+1}}{(k+1+C)!}-\frac{C^k}{(k+C)!}\right)\cdot s_1(k+1,k-R)\\
&=B+\sum_{k=-C}^\infty \frac{C^{k+1}-C^k(k+1+C)}{(k+1+C)!}\cdot s_1(k+1,k-R)\\
&=B-\sum_{k=-C}^\infty \frac{C^k}{(k+1+C)!}\cdot (k+1)s_1(k+1,k-R)\\
&=B-\sum_{k=-C+1}^\infty \frac{C^{k-1}}{(k+C)!}\cdot ks_1(k,k-R-1)\\
&=-\frac{I_{R+1,C}}C
\end{aligned}
$$
with boundary term
$$
\begin{aligned}
B&=\lim_{k\rightarrow\infty}\frac{C^k}{(k+C)!}s_1(k,k-R-1)+\frac{C^{-C}}{(-C+C)!}s_1(-C,-C-R-1)\\
&=\frac{C^{-C-1}}{(-C+C)!}\cdot(-C)s_1(-C,-C-R-1).
\end{aligned}
$$
We can therefore use induction and only have to show $I_{R,C}=0$ for $R=0$. But as Gottfried Helms pointed out, this is easy.
