= Solution
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> gives
$$
\|f_{nm}-x^m\|_\infty\le\sum_{r=0}^{m-1}\frac rn=\frac{m(m-1)}{2n},\qquad |f_{nm}(1)-1|\le\frac{m(m-1)}{2n}.
$$
Using the <supremum norm> contraction in (a) and the <Bernstein falling-factorial identity> in (b),
$$
\begin{aligned}
\|B_n(x^m)-x^m\|_\infty
&\le\|B_n(x^m-f_{nm})\|_\infty+\|B_n(f_{nm})-x^m\|_\infty\\
&\le\|x^m-f_{nm}\|_\infty+|f_{nm}(1)-1|\le\frac{m(m-1)}n.
\end{aligned}
$$
For $m=0,1$, the <monomials> are reproduced exactly. For every fixed $m$, \b[$B_n(x^m)\to x^m$ uniformly on $[0,1]$].
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