We suppose that “$\cdot$” is an associative binary operation of the set $S$.

Let  $f_{1}\!:\,S\to 2^{S}$  be the mapping

We define recursively the mapping

such that

for  $n=2,\,3,\,4,\,\ldots$

For example,

the last equality due to the associativity.  It’s clear that always

We shall show by induction that

for each positive integer $n$.  This means that all groupings of the $n$ fixed elements using parentheses in forming the products with “$\cdot$” yield one single element.

We make the induction hypothesis, that (2) is true for all  $n<k.$

Now let $z,\,z^{{\prime}}$ be arbitrary elements of  $f_{k}(a_{1},\ldots,\,a_{k})$.  Then there exist the elements $x,\,y,\,x^{{\prime}},\,y^{{\prime}}$ of $S$ and the integers  $r,\,s\in\{1,\,\ldots,\,k\!-\!1\}$  such that

If specially  $r=s$,  then, by the induction hypothesis,  $x=x^{{\prime}}$  and  $y=y^{{\prime}}$,  whence  $z=x\cdot y=x^{{\prime}}\cdot y^{{\prime}}=z^{{\prime}}$.  If on the contrary,  $r\neq s$,  e.g.  $r<s$,  then the induction hypothesis guarantees the existence of an element $v$ of $S$ such that

and

Since “$\cdot$” is associative, we have

Thus the equation (2) is in force for  $n=k$.