(combinatorics) 2 problems using signless Stirling number of the first kind 
*

*For every subset of [n-1], take the products of all its elements (empty products being taken to be 1) and then,sum of all 2^(n-1) products. What is this value?


2.For every k-element subsets of [n-1], take the products of all its elements and sum up all n-1Ck products. What is this value?
The formula I've tried is this:
signless Stirling number of the first kind is denoted by c(n,k).
From k=0 to k=n,
∑c(n,k)x^k= x(x+1)...(x+n-1)
In the question number 1,the book says that the products are the coefficients of the polynomial on the right hand side. But, I don't why it is.
In the question number 2, the book says the answer is the coefficient of x^(n-k) on the right hand side. I don't understand it either. And I wonder why I should not change n to n-1 in that formula despite the subsets of [n-1]. 
 A: For  the  second question  we  have  by  inspection that  the  desired
quantity is
$$[x^k] \prod_{q=1}^{n-1} (1+qx).$$
This is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+1}} \prod_{q=1}^{n-1} (1+qz) \; dz.$$
Evaluate this using the negative of the residue at infinity
to get
$$\mathrm{Res}_{z=0}
\frac{1}{z^2} z^{k+1} \prod_{q=1}^{n-1} (1+q/z) 
= \mathrm{Res}_{z=0}
\frac{1}{z^{1-k}}  \prod_{q=1}^{n-1} \frac{z+q}{z}
\\ = \mathrm{Res}_{z=0}
\frac{1}{z^{n-k}}  \prod_{q=1}^{n-1} (z+q)
= \mathrm{Res}_{z=0}
\frac{1}{z^{n-k+1}}  \prod_{q=0}^{n-1} (z+q).$$
This evaluates to
$$\left[ n \atop n-k \right]$$
by inspection.

We then get for the first question
$$\sum_{k=0}^{n-1} \left[ n \atop n-k \right]
= \sum_{k=1}^n \left[ n \atop k \right] = n!$$
which is of course the same as setting $x=1$ in the initial formula.
A: Well, I don't know if it will help you but here is a way to prove it. First look at the products :
$$(a_1+b_1)...(a_{n-1}+b_{n-1})=\sum_{(\epsilon_i)\in\{0,1\}^{n-1}}c_{1,\epsilon_1}...c_{n-1,\epsilon_{n-1}} $$
Where $c_{i,0}=a_i$ and $c_{i,1}=b_i$. Now in our particular case we are looking at :
$$\sum_{k=0}^nc(n,k)x^k=x(x+1)...(x+n-1) $$
Hence, here $a_i=x$ for all $i$ and $b_i=i$ for all $i$. 
$$(x+1)...(x+n-1)=\sum_{(\epsilon_i)\in\{0,1\}^{n-1}}c_{1,\epsilon_1}...c_{n-1,\epsilon_{n-1}}$$
Now we have that if we take $(\epsilon_i)\in\{0,1\}^{n-1}$ and :
$$A:=\{i\in\{1,...,n-1\}|\epsilon_i=1\} $$
Then :
$$c_{1,\epsilon_1}...c_{n-1,\epsilon_{n-1}}=x^{n-1-|A|}\prod_{a\in A}a $$
Hence we have :
$$(x+1)...(x+n-1)=\sum_{(\epsilon_i)\in\{0,1\}^{n-1}}c_{1,\epsilon_1}...c_{n-1,\epsilon_{n-1}}=\sum_{A\subseteq\{1,...,n-1\}}x^{n-1-|A|}\prod_{a\in A}a$$
From this formula, you will get 1 and 2 easily.
