Prove that $1^\alpha + 2^\alpha = 3^\alpha $ if $ \alpha $ is a limit ordinal I am trying to prove the following statement:
Suppose $ \alpha $ is a limit ordinal. Then $ 1 ^\alpha + 2^\alpha = 3^\alpha$.
I'm not sure how to grasp this. Obviously induction won't work, since the statement doesn't hold for non-limit $ \alpha $.
I tried to solve this for $ \alpha = \omega $, but then the statement gets trivial since both sides equal $ \omega $.
I would appreciate a hint

First note that $1^\alpha=1$, always. So really you need to prove that $2^\alpha=3^\alpha$. 
Now you can do this by induction on $\alpha'$ such that $\omega\cdot\alpha'=\alpha$ (and you can also prove there is always a unique such $\alpha'$ for every limit ordinal $\alpha$).
Alternatively, you can prove that $k^{\delta+n}\leq k^\delta+\omega$, when $\delta$ is a limit ordinal and $k,n<\omega$. And if $n=0$, then $k^\delta\leq\delta$. This will give you the wanted result by induction on $\alpha$ being any ordinal.
