Let A+0 = A. If B is the successor of C, let A+B be the successor of A+C. If B is a limit ordinal let A+B be the union of A+C for all C in B.
There is another definition that is equivalent. Take the union of the ordered pairs 0,x and 1,y, for x ∈ A and y ∈ B, and order the set lexicographically. That is, the zero pairs come first, then the one pairs. This is like stacking the elements of B on top of the elements of A. This set is well ordered, hence it is isomorphic to an ordinal. That ordinal is A+B. Verify that A+B has the recursive properties described above.
To find A+B+C, stack the elements of B on top of the elements of A, then stack the elements of C on top of that. This is the same as sorting the ordered pairs 0,x 1,y 2,z where x ∈ A, y ∈ B, and z ∈ C. Clearly (A+B)+C = A+(B+C), hence addition is associative. Note that addition is not commutative, for 1+ω = ω, while ω + 1 = successor(ω). However finite addition is commutative. First note that a finite ordinal cannot be mapped 1-1 onto a smaller ordinal without violating the pigeonhole principle. Then build a bijection between B+A and A+B, and since they are both finite, they must be the same ordinal.
Let A×0 = 0. If B is the successor of C let A×B = A×C+A. If B is a limit ordinal let A×B be the union of A×C for all C in B.
There is another definition that is equivalent. Take the set B cross A and sort the ordered pairs lexicographically. The ordinal associated with this well ordered set becomes A×B. The elements of A are stacked over and over again, once for each entry in B. Adding one more element to B, its successor, pushes another instance of A, hence A×C+A. And when B is a limit ordinal, A×B brings in all the pairs from C cross A for every C in B.
Since three-way multiplication produces the set C cross B cross A, it too is associative. Multiplication is not commutative, since 2×ω = ω, while ω×2 = ω+ω. Finite multiplication is commutative; build a bijection between A cross B and B cross A.
Multiplication left distributes over addition. Both A×B+A×C and A×(B+C) create triples from 0,B,A and 1,C,A, and are sorted the same way to produce the same ordinal. Multiplication does not right distribute over addition, since (1+1)×ω ≠ ω+ω. Multiplication is bi-distributive for finite ordinals; build bijections to show this.
Exponentiation can be defined recursively as A0 = 1, and AB (B successor of C) = AC×A.
Another definition of exponentiation is useful. Think of AB as the set of functions from B into A that are 0 almost everywhere. That is, finitely many members of B are mapped into nonzero members of A. Remember, we don't need the axiom of choice here, because the base sets in the product set are all equal, i.e. B copies of A. If two functions differ then there is a largest value of B on which they differ. If this largest value is z, and one function has f(z) = x while the other has g(z) = y, declare f < g if x < y, and g < f if y < x. Verify that this is a well ordering, hence the ordinal associated with this set is AB. Show this agrees with the inductive definition above.
Show that AB+C = AB×AC.
If all the above were done in rigorous detail it would take up an entire book. In fact Russell and Whitehead (biography) wrote such a book, Principia Mathematica, in three volumes, but few have the patience to read through it and verify all the details. (Yes, Isaac Newton published a seminal book having the same title.)