I have heard about the following result: for each finite simple non-abelian group $S$ and each natural number $r\ge 2$ there exists a number $n=n(r,S)$ such that the power $S^n$ is $r$-generator but $S^{n+1}$ is not $r$-generator. What is known about the numbers $n(r,S)$? Could someone give me references to this, please?


(I have posted this already on mathstackexchange.com, but did not get a response.)


Edit: This question is in a sense a converse to [Bounding from below the cardinality of a set of generators of the $n$-fold cartesian product of a finite group](https://mathoverflow.net/questions/187269/bounding-from-below-the-cardinality-of-a-set-of-generators-of-the-n-fold-carte). There it is basically asked for a given (arbitrary, finite) group $G$ and a given number $n$, how small can a generating set for $G^n$ possibly (not) be. In my question the input parameters were a finite simple group $S$ and a number $r\ge 2$ and the question was how big a number $n$ can possibly be so that $r$ elements are sufficient to generate the power $S^n$. Also I was interested in how this number (the biggest such $n$) is actually computed in concrete examples (or whatever is known about the computation of these numbers). 


Basically I wanted to know, given a finite simple non-abelian group $S$ and a number $r$, the product of how many copies of $S$ do I need to take to get the $r$-generated free object in the formation generated by $S$.


@Editors/moderators: please feel free to delete the question if it is inappropriate.



Powers of finite simple groups
See [Collins's thesis](http://www.math.cornell.edu/m/sites/default/files/imported/Research/SeniorTheses/2010/collinsThesis.pdf), Theorem 2.22, page 21.



> 
> Theorem 2.22. Let $S$ be a nonabelian simple group and
>  $h\_{n-1}(S) < k \le h\_n(S)$. Then $r(S^k)=n$.
> 
> 
> 


Here, $r(G)$ is the minimal number of generators of $G$ and $h\_n(G)$
is the *reduced Euler function* i.e. the number of generator sequences of length $n$ of $G$ divided by $|{\rm Aut}\ G|$. 


