<s>[INST] Power two of ordinal if $\omega$ is ordinal number of set $\mathbb{N}$ can I compute $(\omega + 1)^2$ like below
$\begin{align*}
(\omega + 1)^2 & = \  (\omega + 1).(\omega + 1)\\
& = \ ((\omega + 1).\omega)+(\omega + 1)\\
& = \ (\omega^2 + 1)+(\omega + 1)\\
& = \ \omega ^ 2 + (1 + \omega) + 1\\
& = \ \omega^2 + \omega + 1
\end{align*}$
is $(\omega+1).\omega = \omega^2 + 1$? how can I compute $(\omega+1).\omega$?
 [/INST] Sometimes a picture can help, so I’m adding this addendum to Noah’s answer for the possible benefit of future readers. The first diagram shows the Cartesian product $\omega\times(\omega+1)$; $(\omega+1)\cdot\omega$ is the order type of this product under the lexicographic order.
$$\begin{array}{ccc}
\omega&\color{red}\omega&\color{blue}\omega&\color{green}\omega&\ldots\\
\vdots&\vdots&\vdots&\vdots&\\
4&4&4&4&\ldots\\
3&3&3&3&\ldots\\
2&2&2&2&\ldots\\
1&1&1&1&\ldots\\
0&0&0&0&\ldots
\end{array}$$
Now imagine taking each of the points in the top row and moving it to the bottom of the next column to the right; the result is shown below.
$$\begin{array}{ccc}
\vdots&\vdots&\vdots&\vdots&\vdots\\
5&4&4&4&4&\ldots\\
4&3&3&3&3&\ldots\\
3&2&2&2&2&\ldots\\
2&1&1&1&1&\ldots\\
1&0&0&0&0&\ldots\\
0&\omega&\color{red}\omega&\color{blue}\omega&\color{green}\omega&\ldots\\
\end{array}$$
The lexicographic order on this array is clearly isomorphic to that on the first array, but here it clearly has order type $\omega\cdot\omega=\omega^2$.
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