= Solution
Work on the positive orthant, where the real powers defining $g$ are smooth. Its closure under the constraint is the <compact> weighted simplex $\overline M=\{x_i\ge0,\sum p_ix_i=1\}$, since $0\le x_i\le1/p_i$. The product extends continuously there, is zero on its boundary when $n>1$, and is one at $(1,\ldots,1)$. Thus a <maximum> is attained in the positive interior.
At an interior maximizer the constraint <gradient> $(p_1,\ldots,p_n)$ is nonzero. The multiplier equations give
$$
p_i\frac{g(x)}{x_i}=\lambda p_i\quad\Longrightarrow\quad x_i=g(x)/\lambda\text{ for every }i.
$$
All coordinates are equal, and the weighted constraint forces each to equal one. Therefore the <maximum> is one and $\boxed{g(x)\le1}$ 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.
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