Generalized associativity for a Group

Question

Suppose that we have a Group $(G,.)$, where . is binary operation on $G$. Prove that for any $a_1,a_2,\cdots, a_n\in G, a_1.a_2.a_3.\cdots a_n$ is independent of how it is bracketed.

My attempt: First of all for any $k\in \mathbb N,a_1.a_2.a_3.\cdots a_k$ must be defined in a way. Let's define it like this
$a_1.a_2.a_3.\cdots a_k=a_1.(a_2.(a_3.(\cdots a_k))\cdots) \tag 1$

Now we have to prove that howsoever $a_1.a_2.a_3.\cdots a_n$ is bracketed, it evaluates to form $(1)$, that is $a_1.a_2.a_3.\cdots a_n$ has a unique value which can be given by the form $(1)$.

For $n=1$, $a_1=a_1.e=e.a_1$, where $e$ is identity of the group is a unique element in $G$ as $.$ being a binary operation associates $a_1=a_1.e$ to a unique element in $G$.
For $n=2, a_1.a_2$ is a unique element in $G$ as $.$ is a binary operation on $G$.
For $n=3$, the hypothesis is true by associativity of Group.

Induction hypothesis: Let our hypothesis be true for $r\le n-1$, that is $a_1.a_2.a_3.\cdots a_{r}$ is independent of how it is bracketed and has a unique value which can be given by the form $(1)$.

Now the problem that I'm facing is lies in breaking $a_1.a_2.a_3.\cdots a_{r+1}$ into two parts. I assumed $1\lt i\lt r$ and considered $a_{i+1}.a_{i+2}.a_{i+3}.\cdots a_{r+1}$. As per our definition above,
$a_{i+1}.a_{i+2}.a_{i+3}.\cdots a_{r+1}=a_{i+1}.(a_{i+2}.(a_{i+3}.(\cdots (a_r.a_{r+1}))\cdots )$. RHS is composition of two elements of $G$ under binary operation $.$ and hence unique.
Now $a_1.a_2.a_3.\cdots a_{i}=(a_1.a_2.a_3.\cdots a_{i})$
We have $a_1.a_2.a_3.\cdots a_{i}.(a_{i+1}.a_{i+2}.a_{i+3}.\cdots a_{r+1})=a_1.a_2.a_3.\cdots a_{i}.(a_{i+1}.(a_{i+2}.(a_{i+3}.(\cdots (a_r.a_{r+1}))\cdots )$. Now I'm stuck.

Please see bracket on LHS here, I don't understand how to remove it and once this is removed how do I show that finally the product (compositio under binary operation $.$) evaluates to a unique value given by the form $(1)$.

Please help. Thanks.