Product of $\geq 2^r$ elements can be written as a simple product of $r$ elements? This comes from Jacobson's Basic Algebra I, Exercise 4 of Section 1.4, found on page 42. I don't understand the following problem.

For a given binary composition define a simple product of the sequence of elements $a_1,a_2,\dots,a_n$ inductively as either $a_1u$ where $u$ is a simple product of $a_2,\dots,a_n$ or as $va_n$ where $v$ is a simple product of $a_1,\dots,a_{n-1}$. Show that any product of $\geq 2^r$ elements can be written as a simple product of $r$ elements (which are themselves products).

I thought about approaching this by induction on $r$. But then for $r=1$, it seems to ask that any product of $\geq 2$ elements can be written as a simple product of $1$ element, which sounds unusual to me. The inductive definition given doesn't make sense to me, is there a better way to approach this problem?
 A: This is done by induction (if I understood it correctly). Let us assume that $z$ is a product of at least $2^r$ elements. Because $z$ is a product, it is of the form $z=xy$. Obviously either $x$ or $y$ has at least $2^{r-1}$ factors. By induction hypothesis that factor is a simple product of at least $(r-1)$ elements (most of those products themselves). There $z$ is a simple product of one of the two forms: the shorter of $x$ and $y$ is the $r^{th}$ factor.
The starting point $r=1$ being obvious.
A: The definition is confusing because Jacobson gives here two different definitions at the same time. First one is this:

Simple product of $a_1, a_2, \ldots a_n$ is defined as $a_1u$, where $u$ is a simple product of $a_2, \ldots, a_n$.

The other one is this:

Simple product of $a_1, a_2, \ldots a_n$ is defined as $va_n$, where $v$ is a simple product of $a_1, \ldots, a_{n-1}$.

In the first case the product is of the form $a_1(a_2(a_3(\ldots(a_{n-1}a_n)\ldots)))$ and in the second case it is of the form $(((\ldots(a_1a_2)a_3)\ldots)a_{n-1})a_n$. Both are "simple products" as Jacobson defines them.
Suppose $b_1, b_2, \ldots, b_k$ are elements defined on some binary relation such that $k \geq 2^r$. Let $P$ be some product of these elements, ie. $b_1b_2\ldots b_k$ bracketed in some way. What the question is asking you to show is that there exist elements $a_1, a_2, \ldots a_r$ such that the simple product of $a_1, a_2, \ldots, a_r$ equals $P$.
Now notice that $P = XY$, where $X$ is some product of $b_1, \ldots, b_i$ and $Y$ is some product of $b_{i+1}, \ldots, b_k$. Then either $X$ or $Y$ has $\geq 2^{r-1}$ elements in the product, so by induction one of them is a simple product of $r-1$ elements. In both cases $P$ is a simple product of $r$ elements of one of the two forms.
