Artin's inductive proof of associative law of composition? My question pertains to the inductive proof of associative law of composition quoted here Confused by inductive proof of associative law . 


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*Why $r \leq n-1$? Why did he choose $n-1$?

*Would it be wrong if I prove associativity the following way 
By definition the associative law is valid for $n \leq 2$.
Assume, $[a_{1} ... a_{n}] = [a_{1} ... a_{i}][a_{i+1} ... a_{n}]$ is true for some $n$. 
Then I should show that $[a_{1} ... a_{n+1}] = [a_{1} ... a_{i}][a_{i+1} ... a_{n+1}]$.
So, $[a_{1} ... a_{n+1}] = [a_{1} ... a_{n}][a_{n+1}]$, by definition.
or, $[a_{1} ... a_{n}][a_{n+1}] = ([a_{1} ... a_{i}][a_{i+1} ... a_{n}])[a_{n+1}]$.
By the associative law, $([a_{1} ... a_{i}][a_{i+1} ... a_{n}])[a_{n+1}] = [a_{1} ... a_{i}]([a_{i+1} ... a_{n}][a_{n+1}) = a_{1} ... a_{i}][a_{i+1} ... a_{n+1}]$.
 A: The choice of $n-1$ is merely just a preference to use in induction proofs. It is actually equivalent to choosing $n+1$ instead.
To see why, first note that the inductive step, in general, consists of assuming a proposition is true for some $k$ (or $ n \leq k$) and then showing that $$P(k) \implies P(k+1) \tag{1}$$ If we replace $k$ with $k-1$, we'll get an equivalent statement: 
$$P(k-1) \implies P(k-1+1) =P(k)$$
Therefore instead of proving $(1)$, we can alternatively prove that $$P(k-1) \implies P(k)$$
With that in mind, can you see how your proof is, in essence, the same as the one in the linked question?
I'll present a slightly simpler inductive proof of the same theorem:

In a group, the product of $n \geq 3$ elements $(a_1 \cdot a_2 \cdots a_n)$ does not depend on the arrangement of brackets that define the
  sequence of multiplications. (I.e. $n$-element multiplication is
  associative)

Proof by induction on $n$:
For $n=3$, this is the axiom of associativity in a group. So let $n>3$. Assume, for the purpose of induction, that the proposition is true for multiplicands $<n$. Consider the products $$P=(a_1 \cdot a_2 \cdots a_k)(a_{k+1} \cdots a_n) \space \space \space 1\leq k<n$$ $$Q=(a_1 \cdot a_2 \cdots a_l)(a_{l+1} \cdots a_n) \space \space \space 1\leq l<n$$
For $k=l$ we have that $P=Q$ by the inductive hypothesis. Without loss of generality, let $k<l$. Then by the Inductive hypothesis, we can rewrite $P$ and $Q$ as follows (notice the parentheses): 
$$P=(a_1 \cdot a_2 \cdots a_k)((a_{k+1} \cdots a_l)(a_{l+1} \cdots a_n))$$ $$Q=((a_1 \cdot a_2 \cdots a_k)(a_{k+1} \cdots a_l))(a_{l+1} \cdots a_n)$$
Substituting $a=(a_1 \cdot a_2 \cdots a_k), b = (a_{k+1} \cdots a_l), c= (a_{l+1} \cdots a_n)$ we get $$P=a(bc)$$ $$Q=(ab)c$$
Thus by the group axiom of associativity, $$P=Q \tag{QED}$$
