= Solution
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 $j^{\underline m}\binom nj=n^{\underline m}\binom{n-m}{j-m}$ therefore gives
$$
\begin{aligned}
B_n(f_{nm},x)&=\frac{n^{\underline m}}{n^m}\sum_{j=m}^n\binom{n-m}{j-m}x^j(1-x)^{n-j}\\
&=\frac{n^{\underline m}}{n^m}x^m\sum_{r=0}^{n-m}\binom{n-m}{r}x^r(1-x)^{n-m-r}\\
&=\boxed{f_{nm}(1)x^m}.
\end{aligned}
$$
The 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 $B_n1=1$. This is the <Bernstein falling-factorial identity>.
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