Let $x$ be a real number. I have been trying to simplify the following sum $$\sum_{j=n+1}^{k}\bigg(j^{x-1}-(j-1)^{x-1}\bigg){j\brack j-n+1}$$ for some time now but haven't made any headway. I see that some of those terms will cancel each other, but I have not found any nice way so far to write the result in an explicit closed form without dealing with another summation of the same sort.
Has anyone an idea about how the sum may be simplified (at least in terms of any of the known standard functions)? The notation ${j\brack n}$ denotes the signless Stirling numbers of the first kind.
With telescoping we can simplify the sum at least a bit. We obtain \begin{align*} \color{blue}{\sum_{j={n+1}}^k}&\color{blue}{\left(j^{x-1}-(j-1)^{x-1}\right) \begin{bmatrix} j\\j-n+1\end{bmatrix}}\\ &=\sum_{j={n+1}}^{k}j^{x-1}\begin{bmatrix} j\\j-n+1\end{bmatrix} -\sum_{j=n}^{k-1}j^{x-1}\begin{bmatrix} j+1\\j-n+2\end{bmatrix}\tag{1}\\ &=\sum_{j={n+1}}^{k}j^{x-1}\begin{bmatrix} j\\j-n+1\end{bmatrix} -\sum_{j=n}^{k-1}j^{x-1}\left(j\begin{bmatrix} j\\j-n+2\end{bmatrix} +\begin{bmatrix} j\\j-n+1\end{bmatrix}\right)\tag{2}\\ &=\left(\sum_{j={n+1}}^{k}j^{x-1}\begin{bmatrix} j\\j-n+1\end{bmatrix} -\sum_{j=n}^{k-1}j^{x-1}\begin{bmatrix} j\\j-n+1\end{bmatrix}\right)\\ &\qquad\quad-\sum_{j=n}^{k-1}j^{x}\begin{bmatrix} j\\j-n+2\end{bmatrix}\tag{3}\\ &\,\,\color{blue}{=k^{x-1}\begin{bmatrix}k\\k-n+1\end{bmatrix}-(n-1)! -\sum_{j=n}^{k-1}j^{x}\begin{bmatrix} j\\j-n+2\end{bmatrix}}\tag{4}\\ \end{align*}
Comment:
In (1) we split the sum and shift the index of the right-hand sum by one.
In (2) use the recurrence relation $\begin{bmatrix}p+1\\q\end{bmatrix}=p\begin{bmatrix}p\\q\end{bmatrix}+\begin{bmatrix}p\\q-1\end{bmatrix}$.
In (3) we rearrange the sums.
In (4) do the telescoping noting that $\begin{bmatrix}p\\1\end{bmatrix}=(p-1)!$.