Proof that addition of an ordinal with a non-zero limit ordinal is always a limit ordinal I wanna show that $\alpha + \beta$ is a limit ordinal where $\alpha$ is an ordinal  and $\beta$ is a non-zero limit ordinal.
The less-smart way is to say that it is clearly non-zero since $\beta$ is non-zero and then show that $\alpha + \beta$ is not a successor ordinal, therefore has to be limit ordinal. How can this be proven? Can somebody prove this or perhaps directly prove that it is a limit ordinal?
Secondly, is it true that $\alpha . \beta $ and $\beta . \alpha$ are also limit ordinals? My argument is that using another ordinal $\gamma$, where $\alpha = \gamma + 1$, we can write the multiplications as $\alpha. (\gamma + 1 ) = \alpha . \gamma + \alpha$ and $ (\gamma + 1 ) . \alpha = \gamma . \alpha + \alpha$ respectively. Using the previous part, the results are also limit ordinals.
 A: This seems to follow directly from the definitions, at least as given on Wikipedia in Ordinal arithmetic and Limit ordinal:

...if β is a limit ordinal then α + β is the limit of α + δ for all δ < β.


Addition is strictly increasing and continuous in the right argument:
$$ \delta<\gamma\Rightarrow \alpha+\delta <\alpha+\gamma$$

Hence the set $\{\alpha + \delta \mid \delta < \beta\}$ has no greatest element.

The union of a nonempty set of ordinals that has no greatest element is then always a limit ordinal.

I think for multiplication you meant to write $\beta (\gamma + 1) = \beta \gamma + \beta$. This works if $\alpha$ is a nonzero successor ordinal, but there was no such restriction mentioned (it could be a limit ordinal itself). Also, distribution from the right doesn't work in general, so $(\gamma + 1) \beta$ can't be expanded. However, both these problems can be overcome by essentially the same argument: if $\beta$ is a limit ordinal, then $\alpha \beta$ is defined to be the limit of $\{ \alpha \delta \mid \delta < \beta\}$, and ordinal multiplication is strictly increasing in the right argument.
