 Solution

Consider $B_n$ on the finite-dimensional vector space $E=\mathbb{R}[x{]}_{\le d}$ and use the monomial basis. For $j\ge 1$, a binomial distribution and the falling factorial moment formula givewhere $S(j,\ell )$ is the Stirling number of the second kind and $(n{)}_{\ell }=n(n-1)\cdots (n-\ell +1)$. This follows by expanding $K^j={\sum }_{\ell }S(j,\ell )(K{)}_{\ell }$ and using $\mathbb{E}(K{)}_{\ell }=(n{)}_{\ell }{x}^{\ell }$. Also $B_n(1)=1$.

Each monomial maps to a polynomial of no larger degree. The resulting matrix is triangular with diagonal entries $1$ and $(n{)}_j/n^j$, $1\le j\le d$. When $n\ge d$ and $n\ge 1$, all entries are positive. ThereforeFor constant polynomials the conclusion is immediate. Taking positive indices avoids the undefined sample expression $k/n$ at $n=0$.