Properties of Exponentiation

$x^y = \underbrace{x \cdot x \cdots x \cdot x}_{\text{repeated } y \text{ times}}$

this can be expressed more formally with a recursive definition

$x \cdot x^y = x^{y+1}$

$x^1 = x$

$x^1 = x$ is very intuitive, however, from these simple definitions, one can show that $x^0 = 1$:

$x \cdot x^0 = x^1 = x$

the only multiplicative identity there is (in $\mathbb{R}$) is $1$, so $x^0 = 1$

you can show that $x^a \cdot x^b = x^{a+b}$ easily by using the informal definition:

$x^a \cdot x^b = \overbrace{\underbrace{x \cdot x \cdots x \cdot x}_{\text{repeated } a \text{ times}} \cdot \underbrace{x \cdot x \cdots x \cdot x}_{\text{repeated } b \text{ times}}}^{\text{repeated } a + b \text{ times}} = x^{a+b}$

however, proving it formally is harder

$x^y = x \cdot x^{y-1}$

$x^a \cdot x^b = x^{a-1} \cdot x \cdot x^b = x^{a-1} \cdot x^{b+1}$

inductively,

$x^a \cdot x^b = x^{a-n} \cdot x^{b+n}$

when $n=a$,

$x^a \cdot x^b = x^{a-a} \cdot x^{b+a} = x^0 \cdot x^{b+a} = 1 \cdot x^{b+a} = x^{a+b}$

the final property to be shown is that $(x^a)^b = x^{a \cdot b}$, I actually find this one easier to prove formally than to try to understand it intuitively

$(x^a)^b = x^a \cdot (x^a)^{b-1}$

if we multiply both sides by $x^{an}$,

$x^{an} \cdot (x^a)^b = x^{an} \cdot x^a \cdot (x^a)^{b-1} = x^{an+a} \cdot (x^a)^{b-1} = x^{a(n+1)} \cdot (x^a)^{b-1}$

inductively,

$x^{an} \cdot (x^a)^b = x^{a(n+m)} \cdot (x^a)^{b-m}$

when $n=0$,

when $m=b$,

$(x^a)^b = x^{am} \cdot (x^a)^{b-b} = x^{ab} \cdot (x^a)^0 = x^{ab}$