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 $(\omega+1).\omega$?

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.
