Denote the [unsigned Stirling numbers of the first kind](https://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind) by $s(n,j)$.



> 
> **Question.** Is it true that the polynomials
> $$P\_n(x)=\sum\_{j\geq0}s(n,j)\binom{x}j$$
> have only real roots?
> 
> 
> 


**Note.** Obviously, the roots of $F\_n(x)=\sum\_{j\geq0}s(n,j)\,x^j=\prod\_{k=0}^{n-1}(x+k)$ are all real (integers).



Real-rootedness of some polynomials
Fix $n$ and consider $P\_k(x)=(1+\alpha)(1+2\alpha)\cdots(1+(k-1)\alpha)\cdot{x\choose n}$, where $\alpha$ is the operator sending ${x\choose k}$ to ${x\choose k-1}$. That is, $\alpha$ sends $f(x)$ to $f(x+1)-f(x)$. We'll show by induction that each $P\_k(x)$ has all real roots and that the distance between consecutive roots is at least $1$. The claim is trivial for $k=1$.


Assume the claim holds for $P\_k(x)$. Then $P\_{k+1}(x)=(1+k\alpha)P\_k(x)$ is the linear combination $kP\_k(x+1)-(k-1)P\_k(x)$. Let the roots of $P\_k(x)$ be $a\_1<\cdots<a\_n$. The roots of $P\_k(x+1)$ are $a\_1-1<\cdots<a\_n-1$, which interlace with those of $P\_k(x)$ by the induction assumption. In the notation of [Fisk's "Polynomials, roots, and interlacing"](https://arxiv.org/pdf/math/0612833.pdf), $P\_k(x+1)\underline\ll P\_k(x)$. Let the roots of $P\_{k+1}(x)$ be $b\_1\le\cdots\le b\_n$. It follows from (the proof of) Prop 1.35 that all the roots are real and that $b\_1\le a\_1-1$ and $a\_{i-1}\le b\_i\le a\_i-1$ for all $i>1$. Therefore $b\_i-b\_{i-1}\ge a\_{i-1}-(a\_{i-1}-1)=1$, which completes the induction.


