 Solution

Work on the positive orthant, where the real powers defining $g$ are smooth. Its closure under the constraint is the compact weighted simplex $\overline{M}=\{x_i\ge 0,\sum p_i{x}_i=1\}$, since $0\le x_i\le 1/p_i$. The product extends continuously there, is zero on its boundary when $n>1$, and is one at $(1,\dots ,1)$. Thus a maximum is attained in the positive interior.

At an interior maximizer the constraint gradient $(p_1,\dots ,p_n)$ is nonzero. The multiplier equations giveAll coordinates are equal, and the weighted constraint forces each to equal one. Therefore the maximum is one and $\boxed{g(x)\le 1}$ on $M$. For $n=1$ the constraint itself gives $x_1=1$, so the same conclusion holds directly. Equality occurs exactly at the all-ones point.