I'm wondering whether the following polynomial of a single indeterminate has been studied: take the (partial) Bell polynomial $B_{n,k}$, which is a polynomial in indeterminates $x_1$, $x_2$, …, $x_{n-k+1}$, and replace each indeterminate $x_i$ by the falling factorial $(x)_i=x(x-1) \dots (x-i+1)$. Call this polynomial $N(n,k)$.

I conjecture the following matrix identity. Assemble the polynomials into an infinite lower-triangular matrix $N$. Let $s$ be the lower-triangular matrix of Stirling numbers of the first kind. Let D be the diagonal matrix with entries $x$, $x^2$, $x^3$, ….

Conjecture: $Ns=sD$

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    $\begingroup$ Perhaps the following remark will be useful. One can show by a generating function argument that $$ N(n,k) = \frac{1}{k!}\sum_{i=1}^k(-1)^{k-i} \binom ki\binom{ix}{n}. $$ $\endgroup$ – Richard Stanley Feb 10 '12 at 0:31

As pointed out by Richard Stanley in the comments, from the generating function of the Bell polynomials one finds $$N(n,k) = \frac{1}{k!}\sum_{i=1}^k(-1)^{k-i}\binom ki (ix) _n.$$ Now use the fact that $$(ix) _n=\sum _{j=0}^n s(n,j)i^j x^j,$$ and that $$\frac{1}{k!}\sum _{i=0}^k (-1)^{k-i}\binom{k}{i} i^j=S(j,k),$$ to obtain $$N(n,k)=\sum _{j=k}^n s(n,j)x^jS(j,k).$$ In other words, $N=sDS$, where $s,S$ are the lower triangular matrices of Stirling numbers of first and second kind, respectively. To finish off, notice that $s^{-1}=S$ by Stirling number duality.

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