An ordinal number which satisfies $\omega^{\alpha} = \alpha$

Question

Is there an ordinal such that $\omega^{\alpha} = \alpha$? It seems to me there should be, but I can't explicitly point it out. I know it is possible to prove $\alpha\leq\omega^{\alpha}$, but the equality? Could you give me a helping hand?

Answer 1

Sure there is.

First let's look at $1+\alpha=\alpha$. How can you find an ordinal like that? You start with $1$, and you start adding, $1+1+1+\ldots+1$ until you reach a limit step and things work out.

Now, let's look at $\omega\cdot\alpha=\alpha$. How can you find such an ordinal? You start with $1$, and just take longer and longer products of $\omega$, until you happened across $\omega^\omega$.

So, how would you find $\alpha$ such that $\omega^\alpha=\alpha$?