degree of a polynomial over set-partitions 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
$$f_3(x)=(x)_{t_1+t_2+t_3}-(x)_{t_1+t_2}(x)_{t_3}-(x)_{t_1+t_3}(x)_{t_2}-(x)_{t_2+t_3}(x)_{t_1}+2(x)_{t_1}(x)_{t_2}(x)_{t_3}.$$

Conjecture. The polynomial $f_n$ is of degree $1+(t_1-1)+\cdots+(t_n-1)$ in the variable $x$. Is this true?

 A: 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). 
