what is formula for $\sum_{i=0}^{b-1} (-1)^{i}(b-i)^{m}\binom{b-1}i$ and Real Roots Polynomials Definition 
let $b$ and $m$ are non negative integers with $b\ge 1$
$$D_{m,b}=\sum_{i=0}^{b-1} (-1)^{i}(b-i)^{m}\binom{b-1}i$$
I observed 
$$D_{m,m+1}=m!$$
$$D_{m,m}=(m+1)!/2$$
$$D_{m,m-1}= (m+1)!(3m-2)/24$$
$$D_{m,m-2}= (m+1)!(m-1)(m-2)/48$$
$$\dots$$ 


what is general formula for $D_{m,m-k}$ ?


Let define $F_k=D_{m,m-k}=0$ where $k\in \mathbb{N}$
Example 
$F_1 = 3m-2 = 0  \implies m = 2/3$
$F_2 = (m-1)(m-2) = 0  \implies m = 2,1$


Prove or disprove 
$F_k \implies m \in \mathbb{R} \ \ \forall k$


 A: The formula is 

$$D_{m,b}=(b-1)!\,S(m+1,b)$$ 

where $~S(m,n)~$ are the $\,$ Stirling numbers of the second kind (exist several different notations).
It comes e.g. from $~\displaystyle \sum\limits_{k=0}^\infty\frac{x^k}{k!}D_{m,k+1} = e^{-x}\frac{d}{dx}(e^x \phi_m(x))~$ where $~\phi_n(x)~$ is the Bell polynomial.
This means, that you want to know if all roots of $~S(x+1,x-k)=0~$ are real. 
We can develope the polynomial 

$$S(x,x-k) = \sum\limits_{j=0}^{k-1}c_{k,j}\,{\binom x {\,2k-j~}} ~~ , ~~~~ k\in\mathbb{N}$$

of degree $~2k~$ with the following recursion:
$c_{1,0}:=1 ~~ , ~~ c_{k,j}:=0~$ for $~j\notin\{0,1,…,k-1\} ~~ ,$
$c_{k+1,j} = (k+1-j)\,c_{k,j-1} + (2k+1-j)\,c_{k,j}$ 
Some more values:
$\displaystyle c_{k,0}=\frac{(2k)!}{2^kk!} ~ , ~~ c_{k,1}=\frac{k-1}{3}c_{k,0} ~ , ~~ c_{k,2}=\frac{(k-2)(2k-3)}{6(2k-1)}c_{k,1}$
$\displaystyle c_{k,k-1}=1 ~ , ~~ c_{k,k-2}=2^{k+1}-k-3 ~ , ~~ c_{k,k-3} = \frac{1}{2}(3^{k+2}+k^2+7k+13) – (k+5)2^{k+1}$

Note 1: $~~S(x+1,x-k)=S((x+1),(x+1)-(k+1))$
Note 2: $\displaystyle~\binom {x} {k+1}~$ devides $~S(x,x-k)\hspace{2cm}$ (polynomial division)
$\hspace{1.6cm}$ It remains to solve $\displaystyle~\sum\limits_{j=0}^{k-1}c_{k,j}{\binom {2k-j} {k+1}}^{-1} {\binom {x-k-1} {k-1-j}} =0~$ .
$\underline{\text{Test}}~$ : 
For $~k:=3~$ we get:
$\displaystyle~\sum\limits_{j=0}^{2}c_{3,j}{\binom {6-j} {4}}^{-1} {\binom {x-4} {2-j}} = \frac{(x-2)(x-3)}{2}= 0 ~~~$ => $~~~$ Only real solution values.
For $~k:=4~\,(2<x<5)~$ and $~k:=5~\,(2<x<6)~$ we have also only real solution values.
For $~k:=6~\,(2<x<7)~$ we have three real and two complex solution values.
Or with other words: 
$\hspace{1cm} D_{m,m-5}=0~$ has (all in all) ten real (possibly multiple) and two complex solution values.
And this negates the second question.
A: Actually one of the basic identities for the Stirling N. of 2nd kind is
$$
\left\{ \matrix{
  n \cr 
  m \cr}  \right\}\quad  = {1 \over {m!}}\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,m} \right)} {\left( \matrix{
  m \cr 
  j \cr}  \right)j^{\,n} \left( { - 1} \right)^{\,m - j} }  = {1 \over {m!}}\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,m} \right)} {\left( \matrix{
  m \cr 
  j \cr}  \right)\left( {m - j} \right)^{\,n} \left( { - 1} \right)^{\,j} } 
$$
which confirms the formula already provided by @user90369
$$
D_{\,m,\,\,b}  = \left( {b - 1} \right)!\left\{ \matrix{
  m + 1 \cr 
  b \cr}  \right\}
$$
Now you ask to find the factorization of
$$
D_{\,m,\,\,m - \left( {q - 1} \right)}  = \left( {m - q} \right)!\left\{ \matrix{
  m + 1 \cr 
  m + 1 - q \cr}  \right\}\quad \left| {\;0 \le q \le m} \right.
$$
To this aim,  a possible approach is offered by the very interesting identity 
$$
\left\{ \matrix{
  x \cr 
  x - n \cr}  \right\} = \sum\limits_{0\, \le \,k\, \le \,n} {\left\langle {\left\langle \matrix{
  n \cr 
  k \cr}  \right\rangle } \right\rangle \left( \matrix{
  x + n - 1 - k \cr 
  2n \cr}  \right)} \quad \quad \left| \matrix{
  \;x \in C \hfill \cr 
  \,0 \le {\rm integer }n \hfill \cr}  \right.
$$
which, through the binomial, can be extended also to real and complex $x$.
(re to the renowned "Concrete Mathematics" - pag. 270).
Here the double angle brackets denote the Eulerian N, 2nd kind.
So we have
$$
D_{\,m,\,\,m - \left( {q - 1} \right)}  = \left( {m - q} \right)!\left\{ \matrix{
  m + 1 \cr 
  m + 1 - q \cr}  \right\} = \left( {m - q} \right)!\sum\limits_{0\, \le \,k\, \le \,q} {\left\langle {\left\langle \matrix{
  q \cr 
  k \cr}  \right\rangle } \right\rangle \left( \matrix{
  m + q - k \cr 
  2q \cr}  \right)} \quad \left| {\;0 \le q \le m} \right.
$$
The first values of $q$ give:
$$
\eqalign{
    & \left( {m - 0} \right)!\left\{ \matrix{  m + 1 \cr   m + 1 \cr}  \right\}
 = 1\left( \matrix{  m \cr   0 \cr}  \right)m!
 = m!\quad \left| {\;0 \le m} \right.  \cr 
  & \left( {m - 1} \right)!\left\{ \matrix{  m + 1 \cr   m \cr}  \right\}
 = \left( {m - 1} \right)!\left( {\left\langle {\left\langle \matrix{  1 \cr  0 \cr}  \right\rangle } \right\rangle
 \left( \matrix{  m + 1 \cr   2 \cr}  \right) + \left\langle {\left\langle \matrix{  1 \cr   1 \cr}  \right\rangle } \right\rangle \left( \matrix{  m \cr   2 \cr}  \right)} \right) =   \cr 
  &  = \left( {m - 1} \right)!\left( \matrix{  m + 1 \cr   2 \cr}  \right) = \left( {m + 1} \right)!/2\quad \left| {\;1 \le m} \right.  \cr 
  & \left( {m - 2} \right)!\left\{ \matrix{  m + 1 \cr   m - 1 \cr}  \right\}
  = \left( {m - 2} \right)!\sum\limits_{0\, \le \,k\, \le \,q}
 {\left\langle {\left\langle \matrix{  2 \cr   k \cr}  \right\rangle } \right\rangle \left( \matrix{  m + 2 - k \cr   4 \cr}  \right)}
  = \left( \matrix{  m + 2 \cr   4 \cr}  \right) + 2\left( \matrix{  m + 1 \cr   4 \cr}  \right) =   \cr 
  &  = \left( {m - 2} \right)!{{\left( {m + 2} \right)\left( {m + 1} \right)\left( m \right)\left( {m - 1} \right)
 + 2\left( {m + 1} \right)\left( m \right)\left( {m - 1} \right)\left( {m - 2} \right)} \over {4!}} =   \cr 
  &  = \left( {m - 2} \right)!\left( {m + 1} \right)\left( m \right)\left( {m - 1} \right){{\left( {m + 2} \right) + 2\left( {m - 2} \right)} \over {4!}} =   \cr 
  &  = \left( {m + 1} \right)!{{3m - 2} \over {4!}}\quad \left| {\;2 \le m} \right. \cr} 
$$
as you already found.
To generalize it, let's convert the Binomial into Falling and Rising Factorials
$$
\eqalign{
  & D_{\,m,\,\,m - \left( {q - 1} \right)}  = \left( {m - q} \right)!\left\{ \matrix{  m + 1 \cr   m + 1 - q \cr}  \right\}\quad \left| {\;0 \le q \le m} \right. =   \cr 
  &  = \left( {m - q} \right)!\sum\limits_{0\, \le \,k\, \le \,q} {\left\langle {\left\langle \matrix{  q \cr   k \cr}  \right\rangle } \right\rangle
  \left( \matrix{  m + q - k \cr   2q \cr}  \right)}  =   \cr 
  &  = {{\left( {m - q} \right)!} \over {\left( {2q} \right)!}}\sum\limits_{0\, \le \,k\, \le \,q}
 {\left\langle {\left\langle \matrix{  q \cr   k \cr}  \right\rangle } \right\rangle \left( {m + q - k} \right)^{\,\underline {\,2q\,} } }  =   \cr 
  &  = {{1^{\,\overline {\,m - q\,} } } \over {\left( {2q} \right)!}}\sum\limits_{0\, \le \,k\, \le \,q}
 {\left\langle {\left\langle \matrix{  q \cr   k \cr}  \right\rangle } \right\rangle \left( {m - q + 1 - k} \right)^{\,\overline {\,2q\,} } }  =   \cr 
  &  = {1 \over {\left( {2q} \right)!}}\sum\limits_{0\, \le \,k\, \le \,q} {\left\langle {\left\langle \matrix{  q \cr   k \cr}  \right\rangle } \right\rangle 
{{\Gamma \left( {m - q + 1} \right)\Gamma \left( {m + q + 1 - k} \right)} \over {\Gamma \left( 1 \right)\Gamma \left( {m - q + 1 - k} \right)}}}  =   \cr 
  &  = {1 \over {\left( {2q} \right)!}}\sum\limits_{0\, \le \,k\, \le \,q} {\left\langle {\left\langle \matrix{  q \cr   k \cr}  \right\rangle } \right\rangle
 \left( {m - q} \right)^{\,\underline {\,k\,} } 1^{\,\overline {\,m + q - k} } }  =   \cr 
  &  = \left\{ {\matrix{   {m!} & {0 = q}  \cr 
   {{{\left( {m + 1} \right)!} \over {\left( {2q} \right)!}}\sum\limits_{0\, \le \,k\, \le \,q - 1} {\left\langle {\left\langle \matrix{  q \cr   k \cr}  \right\rangle } \right\rangle
 \left( {m - q} \right)^{\,\underline {\,k\,} } \left( {m + 2} \right)^{\,\overline {\,q - 1 - k} } } } & {1 \le q \le m}  \cr 
 } } \right. \cr} 
$$
the last coming from the fact that the diagonal of the Eulerian Numbers is null, except when both indices are $0$.
In conclusion we can extend $D_{\,m,\,\,m - \left( {q - 1} \right)} $ to complex $m$ as
$$ \bbox[lightyellow] {  
\eqalign{
  & D_{\,z,\,\,z - \left( {q - 1} \right)}  = {{\Gamma \left( {z + 2} \right)} \over {\left( {2q} \right)!}}\sum\limits_{0\, \le \,k\, \le \,q}
 {\left\langle {\left\langle \matrix{  q \cr   k \cr}  \right\rangle } \right\rangle \left( {z - q} \right)^{\,\underline {\,k\,} } \left( {z + 2} \right)^{\,\overline {\,q - 1 - k} } }  =   \cr 
  &  = {1 \over {\left( {2q} \right)!}}\sum\limits_{0\, \le \,k\, \le \,q}
 {\Gamma \left( {z + 1 + q - k} \right)\left\langle {\left\langle \matrix{  q \cr   k \cr}  \right\rangle } \right\rangle \left( {z - q} \right)^{\,\underline {\,k\,} } }
 \quad \left| \matrix{  \;z \in \mathbb C \hfill \cr   \;0 \le q \in \mathbb Z \hfill \cr}  \right. \cr} 
 }$$
where:
 - the case $q=0$ is emcompassed, being
$$
\eqalign{
  & q = 0\quad  \to \,D_{\,z,\,\,z + 1}  = \quad \Gamma \left( {z + 2} \right)\sum\limits_{0\, \le \,k\, \le \,0}
 {\left\langle {\left\langle \matrix{  q \cr   k \cr}  \right\rangle } \right\rangle \left( z \right)^{\,\underline {\,k\,} } \left( {z + 2} \right)^{\,\overline {\, - 1 - k} } }  =   \cr 
  &  = \quad \Gamma \left( {z + 2} \right)\left( {z + 2} \right)^{\,\overline {\, - 1} }  = {{\Gamma \left( {z + 2} \right)} \over {z + 1}} = \Gamma \left( {z + 1} \right) \cr} 
$$
 - the second line comes from considering that the Gamma function does not have zeros, 
yet it has simple poles which annihilate those in the Rising Factorial.
Thus,  concerning the zeros, let's consider first the sum in the first line, alone.
For $1 \le q$, it is a linear combination, with positive integral coefficients,  of $q-1$ polynomials 
each of degree $q-1$ and with integral roots in the interval $\left[ { - q, - 2} \right] \cup \left[ {2,q} \right]$.
None of the roots is common to all the polynomials.
Analyzing the first small values of $q$, it looks that the combination succeeds and preserve for odd $q$ two integral roots 
at $q-2, \; q-1$, and that in any case all the roots have positive real part, so none of them coincides with the poles of Gamma.
So there are $q-1$ roots, some real and some complex.
This might be due to the fact that the Eulerian N. increase at increasing the lower index, but difficult to demonstrate.
