Denote $(x)\_t = x(x-1)(x-2)\cdots(x-t+1)$ and fix some $t\_1,\dots,t\_n\in\mathbb{N}$. Now consider the polynomials
$$f\_n(x)=\sum\_{\pi\in L[n]}(-1)^{\vert\pi\vert-1}(\vert\pi\vert-1)!\prod\_{A\in\pi}(x)\_{\sigma(A)}$$
where the sum extends over all non-empty set-partitions $L[n]$ of $[n]:=\{1,\dots,n\}$ and $\sigma(A)=\sum\_{i\in A}t\_i$.


**Remark.** $\#L[n]=B\_n$ the Bell numbers.


**Example.** Take $n=3$. Then the set of set-partitions of $[3]$ reads 
$$L[3]=\{\{1,2,3\}, \{\{1,2\},\{3\}\}, \{\{1,3\},\{2\}\}, \{\{2,3\},\{1\}\}, \{\{1\},\{2\},\{3\}\}\}$$
and the polynomial becomes
$$f\_3(x)=(x)\_{t\_1+t\_2+t\_3}-(x)\_{t\_1+t\_2}(x)\_{t\_3}-(x)\_{t\_1+t\_3}(x)\_{t\_2}-(x)\_{t\_2+t\_3}(x)\_{t\_1}+2(x)\_{t\_1}(x)\_{t\_2}(x)\_{t\_3}.$$



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> **Conjecture.** The polynomial $f\_n$ is of degree $1+(t\_1-1)+\cdots+(t\_n-1)$ in the variable $x$. Is this true?
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degree of a polynomial over set-partitions
Yes, this is true. Let $M\_1,\dots,M\_n$ be disjoint sets, $|M\_i|=t\_i$, $M=\cup M\_i$. Fix $s\in \{0,1,\dots,n-1\}$. Consider a permutation $w$ of the set $M$ which has exactly $|M|-s$ cycles. Assume that $w$ fixes each of the sets $\cup\_{i\in A} M\_i$, where $A$ is a part of a certain partition $\pi$ of the index set $[n]$. In this case we consider arbitrary cyclic order $F$ on the parts of $\pi$ and take a value $(-1)^{|\pi|-1}$, for each cyclic order, so totally we get $(-1)^{|\pi|-1}(|\pi|-1)!$. I claim that if $s<n-1$, then the sum of all such values equals 0, and if $s=n-1$, it is non-zero. First of all, it is what we have to prove: If you denote $x=-z$ and multiply your polynomial by $(-1)^{\sum t\_i}$, the coefficient of $z^{\sum t\_i-s}$ in the new polynomial $(-1)^{\sum t\_i}f\_n(-z)$ is exactly our sum, this follows from the Stirling numbers interpretations of the coefficients of $z(z+1)\dots (z+t-1)$. 


Note that if $s<n-1$, then any fixed permutation $w$ has a stable set which is a union of several $M\_i$'s (but not all of them). Choose, say, minimal such set of indices $I$ and partition all our values onto pairs by a very natural way: if $I$ was a separate part of $\pi$, unite it with the $F$-next part, else remove $I$ from the part $A$ containing $I$ (such $A$ exists by minimality) and make $I$ an $F$-previous part of a new cyclic order.


If $s=n-1$, this does not work for some permutations (which do not have such stable sets), but for them the sign is always the same (positive). 


