Let $i$, $k$ be integers such that $2 \leq i \leq k$. I would like to show that the sum $$ \sum_{j=1}^{i1} \frac{(1)^{j1}(ij)^k}{(ij)! (j1)!} $$ is positive. I have carried out extensive numerical experiments to check this for small values of $k$. In fact, much more should be true. Define polynomials $$ U(x)=(x+i1)^k $$ and $$ V(x)=x(x+1)\cdots(x+i1). $$ Let $Q$ and $R$ be the quotient and remainder on dividing $U$ by $V$. The above sum is the leading coefficient of $R$. It seems that all the coefficients of $Q$ and $R$ are always positive, and it would be nice to prove this, but I only need the positivity of the above sum. This question has applications for proving the irrationality of certain series.

These are Stirling numbers of the second kind. More precisely your sum is S(k,i1) where $S$ denotes Stirling number of the second kind. 


More general result is the following: if $R$ is remainder of $x^k$ modulo $(xc_1)(xc_2)\dots (xc_i)$ with nonnegative $c_i$'s, then leading coefficient of $R$ is positive. Indeed, let $R=ax^{i1}+\dots$, then $f(x):=x^kax^{i1}\dots$ has roots in $c_i$'s, then by Rolle theorem $f^{(i1)}$ has at least one positive root, which is true iff $a>0$. 

