Show that there exists non-zero countable ordinal $\delta$ such that $\omega\delta=\delta$

Question

I am stuck on the following problem that says:

Show that there exists non-zero countable ordinal $\delta$ such that $\omega\delta=\delta$

My Attempt:

Previously I solved a question similar to this one:

Show that there exists non-zero countable ordinal $\delta$ such that $\omega+\delta=\delta$

where I chose $\delta=\omega^2$, and because of the ordinal arithmetic, $\omega$ on the left is cancelled due to the higher order of $\omega^2$

But, here I couldn't find any idea,

Can someone help me out? Thanks in advance for your time!

Answer 1

Hint: In your previous argument,

\begin{align*} \omega + \omega^2 &= \omega + (\omega +\omega+\dots)\\ & = \omega+\omega+\dots\\ &= \omega^2 \end{align*}

Now try to turn $+$ into $\cdot$ into in this argument. What should you replace $\omega^2$ with?

Better hint: (in response to your comment)

If you replace $+$ with $\cdot$ in the middle equation above, you get: $$\omega\cdot(\omega\cdot\omega\cdot\dots) = \omega\cdot\omega\cdot\dots$$

Now can you see a more compact way of writing the answer?

Answer 2

The two problems are essentially the same intuitively.

In the one you've solved, the mental picture should be $$(\color{red}{\omega})+(\color{green}{\omega+\omega+\omega+\omega+...})=(\color{blue}{\omega+\omega+\omega+\omega+\omega+...}),$$ and the green and blue blocks are the same.

In the new problem, the only difference is that "$+$" has been replaced with "$\cdot$" - so what's the analogous mental picture (and can you turn that into a precise answer/proof)?