Prove that $\xi+\omega= \omega \cdot \xi \ $ iff there is $\zeta$ satisfying $\ \xi=\omega^{\omega}\zeta+1$ I could prove the $\leftarrow$ implication. But assuming $\xi+\omega=\omega\cdot\xi \ $ I couldn't prove that $\xi$ is of the form $\xi=\omega^{\omega}\zeta+1$ for some ordinal $\zeta$.
My attemptive was based on representing $\xi\ $ in its unique cantor normal form, that is $$\xi =\sum\limits_{i=1}^k \omega^{\alpha_i}\cdot m_i$$ with $\alpha_1>\alpha_2>\cdots>\alpha_k$ and $m_i$ finite non-zero ordinals. I tried to use this in $\xi+\omega=\omega\cdot\xi$ and use the uniqueness of the cantor normal form, but I couldn't get anywhere... could you help me with that?

You are on a right track. $\xi+\omega=\omega\cdot\xi$ gives
$$\sum_{i=1}^k \omega^{\alpha_i}\cdot m_i + \omega = \sum_{i=1}^k \omega^{1+\alpha_i}\cdot m_i.$$
Since $1+\alpha_i\ge 1$, $\omega$ is the smallest term of the Cantor normal form. Hence $\omega=\omega^{1+\alpha_k}\cdot m_k$. From this, we have $\alpha_k=0$ and $m_k=1$.
For the remaining terms, we have $\omega^{\alpha_i}\cdot m_i=\omega^{1+\alpha_i}\cdot m_i$. Since each $m_i$ is non-zero, $\omega^{\alpha_i}=\omega^{1+\alpha_i}$. It holds only when $\alpha_i=1+\alpha_i$, which means $\alpha_i=\omega\cdot\beta_i+r_i$ for some $\beta_i\ge 1$ and finite $r_i$.
