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 $(\omega+1).\omega$?
 A: Your answer is right, but your argument is wrong.
Remember that $\alpha\cdot \beta$ means "$\beta$ many copies of $\alpha$" - so e.g. $\omega\cdot 2=\omega+\omega$, whereas $2\cdot\omega=2+2+2+2+...=\omega$.
With this in mind, we have $$(\omega+1)\cdot\omega=\omega+1+\omega+1+\omega+1+...=\omega+(1+\omega)+(1+\omega)+ . . . =\omega^2.$$
So indeed we have $(\omega+1)\cdot(\omega+1)=\omega^2+\omega+1$.
However, it looks like in your argument you assume (third line) that $(\omega+1)\cdot\omega=\omega^2+1$, which is not correct.
A: 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 $\omega\cdot\omega=\omega^2$.
