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At a sampling point $j/n$, the given polynomial has value $f_{nm}(j/n)=n^{-m}{j}^{\underline{m}}$, where $j^{\underline{m}}=j(j-1)\cdots (j-m+1)$ is a falling factorial. It vanishes when $j<m$. The binomial coefficient identity $j^{\underline{m}}\left(\genfrac{}{}{0ex}{}{n}{j}\right)=n^{\underline{m}}\left(\genfrac{}{}{0ex}{}{n-m}{j-m}\right)$ therefore givesThe last step is the binomial theorem, and $n^{\underline{m}}/n^m={\prod }_{r=0}^{m-1}(1-r/n)=f_{nm}(1)$. For $m=0$ the empty product is one and the identity is simply $B_{n}1=1$. This is the Bernstein falling-factorial identity.

Fix $m$ and take $n\ge m$. Comparing the two products defining $f_{nm}$ and the monomial $g_m(x)=x^m$, change one factor at a time. Every factor has absolute value at most one on $[0,1]$, so the triangle inequality givesThe contraction proved in (a) and the Bernstein falling-factorial identity from (b) now giveThis proves uniform convergence for each fixed monomial; linearity then gives it for every fixed polynomial. The cases $m=0,1$ have zero error for every $n$.

Write $j^{\underline{m}}=j(j-1)\cdots (j-m+1)$ for the falling factorial. At the sampling points, $f_{nm}(j/n)=n^{-m}{j}^{\underline{m}}$, which vanishes for $j<m$. For $j\ge m$, cancellation of the factorials givesInsert this into the Bernstein polynomial and set $\ell =j-m$:The binomial theorem evaluates the remaining sum. For $m=0$, both sides equal one. This proves the Bernstein falling-factorial identity $B_n(f_{nm},x)=f_{nm}(1)x^m$.

Fix $m$ and take $n\ge m$. On $[0,1]$, every factor $x-r/n$ and every factor $x$ has absolute value at most one. Replacing the factors successively and applying the triangle inequality givesUsing the supremum norm contraction in (a) and the Bernstein falling-factorial identity in (b),For $m=0,1$, the monomials are reproduced exactly. For every fixed $m$, $B_n(x^m)\to x^m$ uniformly on $[0,1]$.