TITLE: Bounding from below the cardinality of a set of generators of the $n$-fold cartesian product of a finite group
QUESTION [6 upvotes]: Let $G$ be a nontrivial finite group. Given $n\in\mathbb{Z}_{\geq 1}$,
let $G^n$ be the cartesian product of $n$ copies of $G$.
Further let $S\subseteq G^n$ be a generating set of $G^n$.
Question: Do we always have $|S|\geq n$?

REPLY [10 votes]: For an alternative proof of Wiegold's results,
and in particular for finite groups G such that 
the number of generators of G^n is comparable to log n, 
see Section 3 in
Jacques Thévenaz, Maximal subgroups of direct product groups,
Journal of Algebra 198 (1997) 352-361.
Pierre de la Harpe.