Prove $\alpha+\beta=\beta$ if $\alpha < \beta$ $\alpha$ and $\beta$ are ordinal numbers and $\beta$ is limit ordinal.
I assume the proof is relatively simple because it's been omitted in my textbook.
Could someone also provide an intuitive way to think about ordinal arithmetic?
For example (got this from Wikipedia), I can imagine $\omega\times\omega=\omega^2$ to be $1<2<3<...1<2<3<...$ repeating for every natural number. So a countably infinite number of sequences of countably infinite number of numbers. Then it makes sense that $\omega+\omega^2=\omega^2$ because prepending one more sequence of natural numbers doesn't change anything. But how can I generalize this for any ordinal?
Also, this way of thinking doesn't really explain why $\omega^2+\omega$ suddenly isn't equal to $\omega^2$ since the same reasoning could be applied.
 A: The statement is false. For example, $\omega+(\omega+\omega)\neq\omega+\omega$, despite the fact that $\omega<\omega+\omega$.
You can show that $$\alpha+\beta=\beta\iff\alpha\cdot\omega\leq\beta.$$
The proof of the above equality is not difficult:


*

*If $\alpha+\beta=\beta$, then for every $n<\omega$, $\alpha\cdot n+\beta=\beta$. Therefore $\beta\geq\alpha\cdot n$, for all $n<\omega$, so $\beta\geq\sup\{\alpha\cdot n\mid n<\omega\}=\alpha\cdot\omega$.

*In the other direction, if $\alpha\cdot\omega\leq\beta$, then there is some $\gamma$ such that $\alpha\cdot\omega+\gamma=\beta$. Now we have that: $$\alpha+\beta=\alpha+\alpha\cdot\omega+\gamma=\alpha\cdot(1+\omega)+\gamma=\alpha\cdot\omega+\gamma=\beta.$$

An intuitive way of thinking about ordinal addition is that $\alpha+\beta=\gamma$ if we can partition $\gamma$ into an initial segment of $\alpha$ and an end segment of $\beta$ (and nothing else, since it's a partition). 
Now, given two well-orders one is isomorphic to an initial segment of the other, so the difference must be pronounced in the end segments. And this is why $\omega^2+\omega\neq\omega^2$.
