Addition and multiplication on ordinals is defined the same way as it is defined on any other order type. The order types corresponding to sums or products of sequences of ordinals are not necessarily ordinals themselves. For instance, consider the sum Even though the order type $1$ is an ordinal, the sum of a sequence of sets of order type $1$ indexed by $\omega^\ast$ turns out not to be well-ordered. It turns out that it is sufficient for the indexing set to be well-ordered for the sum to exist:
Proposition 1. If $(\alpha_\xi)_{\xi<\beta}$ is an ordinal-indexed sequence of ordinal numbers, then the sum is also an ordinal number.
Proof. Let $(\alpha_\xi){\xi<\beta}$ be an ordinal-indexed sequence of (concrete) ordinals. Then, let $A$ be the set which, by the way, has order type Now, let $A'\subset A$ be nonempty. Since the indexing set is well-ordered, the set has a least element $\xi'$. Additionally, the set $\alpha{\xi'}\times{\xi'}$ is well ordered with the same order type as $\alpha_{\xi'}$, meaning that $A'\cap (\alpha_{\xi'}\times{\xi'})$ has a least element (since it is nonempty by the definition of $\xi'$). This element must be a smallest element of $A'$. For if there were any element in $A'$ smaller than this element, it could not be in $\alpha_{\xi'}\times{\xi'}$, since it was defined as the smallest element in both $A'$ and this set, and it cannot be in any other $\alpha_{\xi}\times{\xi}$, since this would contradict the definition of $\xi'$ as the smallest index of an element appearing in $A'$. Hence, any nonempty subset $A'\subset A$ has a least element, and so $A$ is well-ordered and has the order type of an ordinal. $\blacksquare$
Now, we may define ordinal products $\alpha\cdot\beta$ as "repeated addition", like this:
so that the above proposition guarantees that the product of ordinal numbers is also an ordinal number.
Here are some basic properties of addition on ordinals:
$\beta\leq \alpha+\beta$ for all $\alpha,\beta$.
$\alpha < \alpha+\beta$ for all $\alpha$ and $\beta\ne 0$.
$\gamma+\alpha = \gamma+\beta\implies \alpha = \beta$ for all $\alpha,\beta,\gamma$.
$\alpha\cdot (\beta+\gamma) = \alpha\cdot\beta + \alpha\cdot\gamma$ for all $\alpha,\beta,\gamma$.
$\alpha<\beta\implies \gamma+\alpha<\gamma+\beta$ for all $\alpha,\beta,\gamma$.
Here are some statements that hold for finite numbers, and which might seem at first like they hold for all ordinals, but which in fact do not:
$\alpha+\beta=\beta+\alpha$ fails for many pairs of ordinals, such as $\alpha=1$ and $\beta=\omega$.
$\beta < \alpha+\beta$ for $\beta\ne 0$. This fails for $\alpha=\omega$, $\beta=1$.
$\alpha+\gamma = \beta+\gamma\implies \alpha=\beta$ fails for $\alpha=1,\beta=0,\gamma=\omega$.
$(\alpha+\beta)\cdot\gamma = \alpha\cdot\gamma + \beta\cdot\gamma$ often fails, for instance with $\alpha=\omega,\beta=1,\gamma=\omega$. (Multiplication is left-distributive over addition, but not right-distributive.)
$\alpha<\beta\implies \alpha+\gamma < \beta+\gamma$ fails whenever $\beta>\alpha$ and $\gamma$ is some ordinal large enough to absorb both $\alpha$ and $\beta$. Take, for example, $\alpha=\omega,\beta=\omega^2,\gamma=\omega^3$. (However, this does hold when $\gamma$ is finite.)
Additionally, if $\beta >\alpha$, then there exists a unique ordinal $\gamma$ such that $\beta=\alpha+\gamma$. This means that we may define subtraction on ordinals as follows: if $\beta>\alpha$, let $(\beta-\alpha)$ be the unique ordinal such that $\beta = \alpha+(\beta-\alpha)$.
Here is a useful property of ordinal multiplication and its interaction with strict inequality:
Proposition 2. If $\alpha,\beta,\gamma>0$ are ordinals, then $\gamma\cdot\alpha < \gamma\cdot\beta$ $\iff$ $\alpha < \beta$.
Proof. Suppose $\alpha<\beta$ so that $\beta=\alpha+\delta$ for some $\delta>0$. Then we have that $\gamma\cdot\beta=\gamma\cdot(\alpha+\delta)=\gamma\cdot\alpha+\gamma\cdot\delta$, and $\gamma\cdot\delta>0$ since $\gamma,\delta>0$, meaning that $\gamma\cdot\alpha<\gamma\cdot\beta$. Thus, $\alpha<\beta\implies\gamma\cdot\alpha<\gamma\cdot\beta$.
The inverse is straightforward. Suppose that $\alpha\geq\beta$, so that either $\alpha=\beta$, in which case $\gamma\cdot\alpha=\gamma\cdot\beta$, or $\alpha>\beta$, in which case $\gamma\cdot\alpha>\gamma\cdot\beta$ by the above reasoning. In either case, $\gamma\cdot\alpha<\gamma\cdot\beta$ fails to hold. Thus, $\alpha\not<\beta\implies\gamma\cdot\alpha\not<\gamma\cdot\beta$, and therefore $\gamma\cdot\alpha<\gamma\cdot\beta\implies\alpha<\beta$ as desired. $\blacksquare$
If $\alpha+\beta=\beta$, we say that $\beta$ absorbs $\alpha$. The following proposition sheds light on when exactly one ordinal is "big enough" to absorb another:
Proposition 3. $\beta$ absorbs $\alpha$ iff $\beta\geq \alpha\cdot\omega$.
Proof. Notice that if $\beta$ absorbs $\alpha$, then all ordinals greater than $\beta$ also absorb $\alpha$, for any ordinal greater than $\beta$ can be written in the form $\beta+\gamma$, and we have $\alpha+(\beta+\gamma)=(\alpha+\beta)+\gamma=\beta+\gamma$. So it suffices to show that $\alpha\cdot\omega$ is the smallest ordinal absorbing $\alpha$.
Clearly $\alpha\cdot n$ does not absorb $\alpha$ for any finite ordinal $n$, for $\alpha+\alpha\cdot n = \alpha\cdot (n+1) = \alpha\cdot n + \alpha > \alpha\cdot n$. However, $\alpha\cdot\omega$ does absorb $\alpha$, since $\alpha+\alpha\cdot\omega = \alpha\cdot(1+\omega)=\alpha\cdot\omega$. For any ordinal less than $\alpha\cdot\omega$, there is some ordinal $\alpha\cdot n$ greater than it which does not absorb $\alpha$, meaning that $\alpha\cdot\omega$ is the smallest ordinal absorbing $\alpha$, and the ordinals that absorb $\alpha$ are precisely those greater than $\alpha\cdot\omega$. $\blacksquare$
The following fact is helpful when calculating finite powers of ordinals, particularly utilizing their unique decomposition into prime components :
Proposition 4. If $\beta$ absorbs $\alpha$, then for any $n\geq 2$, if $\alpha$ is a successor ordinal, we have and if $\alpha$ is a limit ordinal, we have
Proof. First we tackle the first statement and prove it by induction. Suppose that $\alpha$ is a successor ordinal, and that the statement is true for some $n\in\mathbb N$, so that Now notice that $\beta^n$ necessarily absorbs $(\beta^{n-1}+\cdots+1)\alpha$, which means that meaning that the claimed formula is also true for case $n+1$. Since it holds true trivially for $n=1$, giving $(\beta+\alpha)^1 = \beta+\alpha$, we have that it holds for all $n\in\mathbb N$ by induction.
Now suppose $\alpha$ is a limit ordinal. If we assume that $(\beta+\alpha)^n=\beta^n+\beta^{n-1}\alpha$ for some $n\in\mathbb N$, then we have that $\beta^n$ absorbs $\beta^{n-1}\alpha$, and therefore and the claimed formula also holds for $n+1$. Since it is again trivially true for $n=1$, the formula holds in general by induction. $\blacksquare$