In a finite field $F$, say of order 7. Suppose I didn't known that it must isomorphic to $\mathbb{F}_7$
By the definition of field, there exists $1 \in F$, and hence exists $1+1$ (denoted by $2$), $1+1+1$(denoted by $3$), ... ,6, and 0.
Then why $2*3$ must be $6$ in $F$, $\{1, 2, 3, 4, 5, 6, 0 \}$ is isomorphic to $\mathbb{F}_7$ as an additive group clearly, but not so clear an isomorphism as a field.
Why the operation in the field is the same as usual in general?
$6$ denotes the element $1+1+1+1+1+1$. Now $(1+1)(1+1+1)=1+1+\cdots1 (6 times)$ by expanding the product (distributive property).