As we know, $2^{\aleph\_0}$ is a cardinal number, so it is a limit ordinal number. However, it must not be $2^\omega$, since $2^\omega=\sup\{2^\alpha|\alpha<\omega\}=\omega=\aleph\_0<2^{\aleph\_0}$, and even not be $\sum\_{i = n<\omega}^{0}\omega^i\cdot a\_i$ where $\forall i \le n[a\_i \in \omega]$. Since $\|\sum\_{i = n<\omega}^{0}\omega^i\cdot a\_i\| \le \aleph\_0$ for all of them.

Besides, $\sup\{\sum\_{i = n<\omega}^{0}\omega^i\cdot a\_i|\forall i \le n(a\_i \in \omega)\}=\omega^\omega$, and $\|\omega^\omega\|=2^{\aleph\_0}$ since every element in there can be wrote as $\sum\_{i = n<\omega}^{0}\omega^i\cdot a\_i$ where $\forall i \le n[a\_i \in \omega]$ and actually $\aleph\_{0}^{\aleph\_0}=2^{\aleph\_0}$ many. 

Therefore $\omega^\omega$ is the least ordinal number such that has cardinality $2^{\aleph\_0}$, and all ordinal numbers below it has at most cardinality $\aleph\_0$. Hence $\omega^\omega=2^{\aleph\_0}=\aleph\_1$?