I see what you're trying to do here, but the wording needs some work.
I actually proved this in an assignment once. It was a great work of passive agression against my lecturer who insisted vehemently that no steps be missed. So, I decided to try formally proving generalised associativity (and commutativity too). Here's a cleaned-up version of the proof I gave him.
Lemma: (Generalised Associativity) Suppose $(G, *)$ is a group. We can recursively define a sequence $(*^n : G^n \rightarrow G)_{n=1}^\infty$ by:
$$*^1 : x \mapsto x$$
$$*^{n+1} : (x_1, \ldots , x_{n+1}) \mapsto (*^n(x_1, \ldots , x_n)) * x_{n+1}$$
Then $*^2 = *$ and, for any $m, n \in \mathbb{N}$, and $(x_1, \ldots, x_{m+n}) \in G^{m+n}$, we have:
\begin{equation}
*^{m+n}(x_1, \ldots, x_{m+n}) = (*^m(x_1, \ldots, x_m)) * (*^n(x_{m+1}, \ldots, x_{m+n})) \tag{$\star$}
\end{equation}
Proof.
The first claim that $*^2 = *$ follows immediately from the recurrence relation with $n = 1$. To prove the second claim, we proceed by induction on $n$. That is, for each $n \in \mathbb{N}$, we wish to prove ($\star$) for all $m \in \mathbb{N}$. For $n = 1$, the equality to be proven is simply the recurrence relation, so we have our base case. Fix $n > 1$, and suppose the claim is true for $n - 1$ instead of $n$. Then, for any $(x_1, \ldots, x_{m + n}) \in G^{m + n}$, we have:
\begin{align*}
*^{m + n}(x_1, \ldots, x_{m+n}) &= (*^{m+n-1}(x_1, \ldots, x_{m+n-1})) * x_{m+n} \\
&= ((*^m(x_1, \ldots, x_m)) * (*^{n-1}(x_{m+1}, \ldots, x_{m+n-1}))) * x_{m+n} \\
&= (*^m(x_1, \ldots, x_m)) * ((*^{n-1}(x_{m+1}, \ldots, x_{m+n-1})) * x_{m+n}) \\
&= (*^m(x_1, \ldots, x_m)) * (*^n(x_{m+1}, \ldots, x_{m+n}))
\end{align*}
by the recurrence relation, inductive hypothesis, associativity, and the recurrence relation respectively. By induction, we have proven the claim. QED.
The notation of generalised associativity is rather painful, but it lays the foundations for the common, and far more elegant notation for group operations (and associative operations in general). We typically denote:
$$x_1 * x_2 * \ldots * x_n := *^n (x_1, x_2, \ldots, x_n)$$
Notice that, because $*^2 = *$, this notation is consistent with the infix notation of the binary operation $*$. The equality ($\star$) now translates to the far less intimidating:
$$x_1 * \ldots * x_m * x_{m+1} * \ldots * x_{m+n} = (x_1 * \ldots * x_m) * (x_{m+1} * \ldots * x_{m+n})$$
In essence, this means we can consistently write strings of group elements delineated by the group operation, and place parentheses anywhere (so long as the formula remains well-formed) without affecting the overall result of the operation.
All of that said, I do realise that the exposition at the end is a bit of a cop-out. In order to do things properly, you need to be able to encode the idea of placing brackets in order to form a well-formed expression. My suggestion would be to form a parse tree, and prune it inductively.
Anyway, I hope this helps in some way.
Best Answer
You are right.
I wouldn't try to evaluate parts of it, such as looking at the value of $(\ldots * b_k)$, but instead consider it a string of symbols. There are two repeating patterns, on either side of $b_k$: $$\underbrace{\underbrace{b_1 \, * \, (}\, \underbrace{b_2 \, * \, (} \, \underbrace{b_3 \, * \, (} \ldots \underbrace{b_{k-1} \, * \, (}}_{\text{A pattern of $k-1$ "blocks"}} \, b_k \, \underbrace{)))\ldots)}_{\text{$k - 1$ times}}$$ Note, in my formulation, I included some extra superfluous parentheses around $b_k$, just to emphasise the pattern. But, other than these extra parentheses, I am saying the same thing as Dummitt and Foote.