Why do you need induction to show that $x^{a+b}=x^ax^b$ if $x\in G$ I have seen a couple of proofs online for problem #1.1.19 in Dummit and Foote's Abstract Algebra that use strong induction to prove the following claim;

Let $x\in G$ where $G$ is a group.  Let $a,b\in \mathbb{Z}^+$.  Prove $$x^{a+b}=x^ax^b$$

I have not seen any proofs of the following form:
$$x^{a+b}=\underbrace{x\cdot x\cdots x}_{a+b\text{ times}}=\underbrace{x\cdot x\cdots x}_{a\text{ times}}\cdot \underbrace{x\cdot x\cdots x}_{b\text{ times}}=x^ax^b$$
I would assume that with positive integer exponents, this would be the most obvious way to show this is true.  So why the need for induction?
 A: Note that multiplication is defined as a binary operator, not as something you do with a bunch of copies of $x$.  As such, when you declare that 
$$
\underbrace{x\cdot x\cdots x}_{a+b\text{ times}}=\underbrace{x\cdot x\cdots x}_{a\text{ times}}\cdot \underbrace{x\cdot x\cdots x}_{b\text{ times}}
$$
you're assuming that the only piece of information that matters is the number of elements x which are being multiplied. That is, you're implicitly moving around a lot of parentheses in the equation
$$
\underbrace{(((x\cdot x)\cdots )x)}_{a+b\text{ times}}=\underbrace{(((x\cdot x)\cdots )x)}_{a\text{ times}}\cdot \underbrace{(((x\cdot x)\cdots )x)}_{b\text{ times}}
$$
You'll notice that the inductive step  in the proof only requires that we deal with one reordering at a time, as in
$$
x^{a+1} \cdot x^b = (x \cdot x^a) \cdot x^b = x \cdot (x^a \cdot x^b) = x \cdot x^{a + b} = x^{a + b + 1}
$$
A: Because, it's considered too informal. To be formal, one needs to use the definition of $x^n$, and the definition is not the product of $n$ copies of $x$.

Since for nonnegative integers $n$, the definition of $x^n$ is an inductive definition, a formal proof of that identity is pretty much forced to use induction.
