I fix $n\in\mathbf{N}^{*}$ and $n$ elements $\alpha_1,\ldots,\alpha_n$ in $\mathbf{N}^{*}$. Consider the polynomial $$Q(T)=\prod\limits_{1\leq i \leq n} \prod\limits_{0\leq j \leq \alpha_i -1} (\alpha_i T + \alpha_i - j)$$ from $\mathbf{Z}[T]$. This polynomial takes obvioulsy integral values on the integers so that a well-known theorem tells us that we can write $Q$ as $\mathbf{Z}$-linear combination of the polynomials ${T \choose k} = \frac{T(T-1)\ldots (T-k+1)}{k!}$ for $k\in\{0,\ldots,d\}$ (so that ${T \choose 0} = 1$) where $d = \sum\limits_{1\leq i \leq n} \alpha_i$ is the degree of $Q$.
Remark. Any $\mathbf{Z}$-linear combination of this kind takes integral values on the integers, and in fact any polynomial of $\mathbf{Q}[T]$ taking integral values on the integers has a unique such decomposition, this statement being the aforementioned theorem. For a reference, see the paragraph about the Hilbert-Serre's theorem in Hartshorne's Algebraic Geometry.
So let's write $Q(T) = \sum\limits_{1\leq k \leq d} c_k {T \choose k}$ with the $c_k$'s in $\mathbf{Z}$. A straightforward calculation shows us that if we set $b_k := \frac{c_k}{\alpha_1 ! \ldots \alpha_n !}$ for $k\in\{0,\ldots,d\}$ we have $$b_k = \sum\limits_{0\leq l \leq k} (-1)^{k+l} {k \choose l} \prod\limits_{1\leq i \leq n} {(l+1)\alpha_i \choose \alpha_i}$$ for $k\in\{0,\ldots,d\}$.
My question is the following : can we find a prime $p$ dividing all the $b_k$'s ?
I conjecture/feel that we can't, but cannot prove it. Help would be greatly appreciated.