Variation on Stirling numbers of the first kind - What properties do they have? According to this paper by Daniel Grunberg and Wikipedia, the Stirling numbers of the first kind are related to the generalized Harmonic numbers.  
For instance
$\begin{bmatrix}n+1\\0+1\end{bmatrix}=n!$
$\begin{bmatrix}n+1\\1+1\end{bmatrix}=n!H_n$
$\begin{bmatrix}n+1\\2+1\end{bmatrix}=n!\left[\frac{1}{2}(H_n)^2-\frac{1}{2}H_n^{(2)}\right]$
$\begin{bmatrix}n+1\\3+1\end{bmatrix}=n!\left[\frac{1}{6}(H_n)^3-\frac{1}{2}H_nH_n^{(2)}+\frac{1}{3}H_n^{(3)}\right]$
$\begin{bmatrix}n+1\\4+1\end{bmatrix}=n!\left[\frac{1}{24}(H_n)^4-\frac{1}{4}(H_n)^2H_n^{(2)}+\frac{1}{8}(H_n^{(2)})^2+\frac{1}{3}H_nH_n^{(3)}-\frac{1}{4}H_n^{(4)}\right]$
$\ \ \ \vdots$
For signed Stirling numbers we have the relation
$\begin{bmatrix}n\\i\end{bmatrix}=(-1)^{n-i}s(n,i)$
These identities are defined via the partitions of $i$ in $\begin{bmatrix}k\\i+1\end{bmatrix}$ (see links above).
There are many fomulas that use the properties of the Stirling numbers of the first kind.  I am interested in exploring the properties of the Stirling-type function that we get when we replace the generalized Harmonic number with a sum that runs over any set of numbers (not just the natural numbers), raised to the power $s$. 
So for a set $A=\{a_1,a_2,a_3,...,a_n\}$ we will define
${\begin{bmatrix}n\\i\end{bmatrix}}_{A^s}$ to be the Stirling-type number that is related to $$f_A(s)=\sum_{k=1}^n\frac{1}{{a_k}^s}$$ in the same way that the regular Stirling number is related to the generalized Harmonic number.  There is no reason (as far as I know) to still have the $+1$s in our new Stirling-type number relation so for simplicity we'll discard those taking ${\begin{bmatrix}n+1\\i+1\end{bmatrix}}\to{\begin{bmatrix}n\\i\end{bmatrix}}_{A^s}$ in our analogy so now we have these relations:
$\begin{bmatrix}n\\0\end{bmatrix}_{A^s}=\prod_{k=1}^n {a_k}^s$
$\begin{bmatrix}n\\1\end{bmatrix}_{A^s}=\left(\prod_{k=1}^n {a_k}^s\right)f_A(s)$
$\begin{bmatrix}n\\2\end{bmatrix}_{A^s}=\left(\prod_{k=1}^n {a_k}^s\right)\left[\frac{1}{2}(f_A(s))^2-\frac{1}{2}f_A(2s)\right]$
$\begin{bmatrix}n\\3\end{bmatrix}_{A^s}=\left(\prod_{k=1}^n {a_k}^s\right)\left[\frac{1}{6}(f_A(s))^3-\frac{1}{2}f_A(s)f_A(2s)+\frac{1}{3}f_A(3s)\right]$
$\begin{bmatrix}n\\4\end{bmatrix}_{A^s}=\left(\prod_{k=1}^n {a_k}^s\right)\left[\frac{1}{24}(f_A(s))^4-\frac{1}{4}(f_A(s))^2f_A(2s)+\frac{1}{8}(f_A(2s))^2+\frac{1}{3}f_A(s)f_A(3s)-\frac{1}{4}f_A(4s)\right]$
$\ \ \vdots$
So what properties can we prove about these Stirling-type numbers? Concerning their:
$\bullet$ recurrence relations (if any)?
$\bullet$ generating function?
$\bullet$ simple identities? I've discovered these
$$\begin{align}
&\begin{bmatrix}n\\n\end{bmatrix}_{A^s}=1\\
&\begin{bmatrix}n\\n-1\end{bmatrix}_{A^s}=\sum_{k=1}^n {a_k}^s\\
&\begin{bmatrix}n\\n-2\end{bmatrix}_{A^s}=\sum_{k=1}^{n-1}\left({a_k}^s\sum_{j=k+1}^n{a_j}^s\right)\\
&\begin{bmatrix}n\\n-3\end{bmatrix}_{A^s}=\sum_{k=1}^{n-2}\left({a_k}^s\sum_{j=k+1}^{n-1}\left({a_j}^s\sum_{i=j+1}^n{a_i}^s\right)\right)
\end{align}$$
And this pattern continues.
$\bullet$ sum or product identities? I've also discovered that
$$\frac{1}{\prod_{k=1}^n{a_k}^s}\sum_{i=0}^n (-1)^{i}\begin{bmatrix}n\\i\end{bmatrix}_{A^s}=\prod_{k=1}^n\left(1-\frac{1}{{a_k}^s}\right)$$
which is interesting because it is the reciprocal of the zeta function when we input primes and take $n\to\infty$.
A further simple question I have would be, what uses for combinatorics (or other fields of math) does this have?
Any references to this topic are gladly welcomed.
 A: In defining these varied Stirling numbers by their ordinary generating function, many of the identities can be shown to be equivalent. Let $ A = \{ a_0, a_1, \cdots, a_\infty\}$ and define:
$$
\begin{bmatrix}n\\k\end{bmatrix}_A = \left [x^k \right] x^{\overline{n}}_A = 
\left [x^k \right] \prod_{p = 0}^{n-1} (x + a_p)
$$
Where $ \left [x^k \right] $ denotes the coefficient of $ x^k $ in the power series expansion. The reason for the $ +1 $s in the original definition come from the fact that the normal Stirling numbers are defined over the set $ \{ 0, 1, \cdots, n - 1 \} $ giving the total of $ n $ numbers, but, to get to $ n $ in the formula, the $ +1 $s would be required.
Recurrence Formula:
$$
\begin{bmatrix}n+1\\k\end{bmatrix}_A = a_n\begin{bmatrix}n\\k\end{bmatrix}_A + \begin{bmatrix}n\\k-1\end{bmatrix}_A
$$
Proof - By the generating function:
$$
x^{\overline{n+1}}_A = x^{\overline{n}}_A (x + a_n) = x \ x^{\overline{n}}_A + n x^{\overline{n}}_A
$$
The coefficient of $ x^k $ on the far left is $ \begin{bmatrix}n+1\\k\end{bmatrix}_A $ , while on the right side it is $ \begin{bmatrix}n\\k-1\end{bmatrix}_A + a_n\begin{bmatrix}n\\k\end{bmatrix}_A $ , therefore the recurrence holds. $ \tag*{$\blacksquare$} $
Simple Identities:
The $ k $th elementary symmetric polynomial may be defined as:
$$
e_k(a_0, \cdots , a_{n-1}) = \left [ x^{n-k} \right ] \prod_{p = 0}^{n-1} (x + a_p) = \begin{bmatrix}n\\n-k\end{bmatrix}_A
$$
These given sums are then variations on the normal $ k $-tuple sum:
$$
e_k(a_0, \cdots , a_{n-1}) = \sum_{0 \le j_1 \le \cdots \le j_k \le n-1} a_{j_1} \cdots a_{j_n}
$$ $ \tag*{$\blacksquare$} $
Generalized Harmonic Numbers:
Let $ A_n! = \prod_{p=0}^{n-1} a_p $. Notice that for $ 0 \le k \le n-1 $ and nonzero $ a_k $:
$$
e_{n-k} (a_0, \cdots , a_{n-1}) = A_n! \ e_k \left (\frac{1}{a_0}, \cdots , \frac{1}{a_{n-1}} \right)
$$
By the Girard-Waring formulae, as in here:
$$
e_k(\cdots) = \sum \prod_{i=1}^n \frac{(-1)^{m_i}}{m_i!} \left ( \frac{p_{i} (\cdots)}{i} \right )^{m_i}
$$
Where the sum is over all non-negative $ m_i $ such that $ m_1 + 2 \ m_2 + \cdots + n \ m_n = n $. Now using the fact that:
$$
p_k \left (\frac{1}{a_0}, \cdots , \frac{1}{a_{n-1}} \right) = \sum_{s = 0}^{n-1} \frac{1}{a_s^k} = H^{(k)}_{A_n}
$$
Putting this all together gives exactly the forms from your links:
$$
\begin{bmatrix}n+1\\k+1\end{bmatrix}_A = e_{n-k} (a_0, \cdots , a_n) = A_{n+1}! e_{k}\left (\frac{1}{a_0}, \cdots , \frac{1}{a_n} \right) = \sum \prod_{i=1}^n \frac{(-1)^{m_i}}{m_i!} \left ( \frac{H^{(i)}_{A_{n+1}}}{i} \right )^{m_i}
$$ $ \tag*{$\blacksquare$} $
Lastly your sum-product identity is obtained from the generating function when $ x=-1 $ and then dividing through by $ A_n! $. It does give the form of a generic Euler Product, and, with a judicious choice of parameters, could give rise to many L-Functions. However, studying their properties may not be as easy to do with the Stirling approach as it is with their normal routes.
A: Just to complement Paul's answer, which already well put the picture in frame, let me add that the product $\prod_{p = 0}^{n-1} (x + a_p)$ can also be developed by Vieta's formulas, which give a straightforward expression for $\begin{bmatrix}n\\k\end{bmatrix}_A $.
Also consider to define the recurrence 
$$
\left\{ \begin{gathered}
  n + 1 \\ 
  m \\ 
\end{gathered}  \right\}_{\;A}  = a_{\,m} \left\{ \begin{gathered}
  n \\ 
  m \\ 
\end{gathered}  \right\}_{\;A}  + \left\{ \begin{gathered}
  n \\ 
  m - 1 \\ 
\end{gathered}  \right\}_{\;A} \quad \left| {\;\left\{ \begin{gathered}
  0 \\ 
  m \\ 
\end{gathered}  \right\}_{\;A}  = \delta _{\,m,\,0} } \right.
$$
and you will get
$$
x^{\,n}  = x_{\left\{ {0,0, \cdots } \right\}} ^{\,\overline {\,n\,} }  = \sum\limits_k {\left( { - 1} \right)^{\,n - k} \left\{ \begin{gathered}
  n \\ 
  k \\ 
\end{gathered}  \right\}_A x_A ^{\,\overline {\,k\,} } }  = \sum\limits_k {\left\{ \begin{gathered}
  n \\ 
  k \\ 
\end{gathered}  \right\}_A x_A ^{\,\underline {\,k\,} } } 
$$
So it is clear that a cascade of further developments is possible.
