= Solution
Integrating the <uniform density> gives $\mathbb EX^k=\int_0^1x^k\,dx=1/(k+1)$. By <independence of random variables>, $\mathbb E(XY)^k=\mathbb EX^k\,\mathbb EY^k=1/(k+1)^2$. For the remaining <expected value>, integrate first in $x$:
$$
\begin{aligned}
\mathbb E(1-XY)^k
&=\int_0^1\int_0^1(1-xy)^k\,dx\,dy\\
&=\frac1{k+1}\int_0^1\frac{1-(1-y)^{k+1}}y\,dy
=\frac1{k+1}\sum_{j=0}^k\int_0^1(1-y)^j\,dy.
\end{aligned}
$$
The value at $y=0$ is supplied by continuity. Therefore, with the <harmonic number> $H_m=\sum_{j=1}^m1/j$,
$$
\boxed{\mathbb E(1-XY)^k=\frac{H_{k+1}}{k+1}.}
$$
Conditional on one sample point being $(x,y)$, it is a <maximal external point> precisely when none of the other $n-1$ points falls in the upper-right rectangle $(x,1]\times(y,1]$. Its area is $(1-x)(1-y)$, so its conditional probability of being maximal is $[1-(1-x)(1-y)]^{n-1}$. Both $1-X$ and $1-Y$ are independent <uniform random variables>, hence a given point is maximal with probability $H_n/n$. Summing their <indicator random variables> and applying <linearity of expectation> gives the <expected number of coordinatewise maxima>:
$$
\boxed{\mathbb E[\text{number of maximal external points}]=H_n.}
$$
This grows only logarithmically with the sample size, even though all $n$ points are potential candidates.
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