 Bernstein falling-factorial identity

At the sampling point $j/n$, the polynomial inside the Bernstein polynomial has value $n^{-m}j(j-1)\cdots (j-m+1)$. The falling factorial identity $j^{\underline{m}}\left(\genfrac{}{}{0ex}{}{n}{j}\right)=n^{\underline{m}}\left(\genfrac{}{}{0ex}{}{n-m}{j-m}\right)$ and the binomial theorem prove the displayed formula. For fixed $m$, both the input and its scalar factor tend uniformly to $x^m$ as $n$ grows. The supremum norm contraction of the Bernstein polynomial therefore gives $\Vert B_n(x^m)-x^m{\Vert }_{\infty }\le m(m-1)/n$ for $n\ge m$.