How do you best simplify $(\omega + 1) \cdot \omega^{2}$? I am currently reading "An Introduction to Set Theory" by Hrbacek and Jech. I have come to an exercise that I am having trouble with. The exercise asks the reader to simplify three ordinal arithmetic expressions. I am having trouble simplifying the last expression.
The last expression is given below :
\begin{equation}
(\omega + 1) \cdot \omega^{2}
\end{equation}
Here $\omega$ represents the natural numbers (the smallest limit ordinal).
I am not sure how to simplify this, my attempt so far is given below :
\begin{align}
(\omega + 1) \cdot \omega^{2} 
 & = (\omega + 1) \cdot (\omega^{2}) \\
 & = (\omega + 1) \cdot (\omega \cdot \omega)\\
 & = \left( (\omega + 1) \cdot \omega \right) \cdot \omega \\
 & = \left( \sup\left( \{(\omega + 1) \cdot n \; \mid \; n \in \boldsymbol{N} \} \right) \right) \cdot \omega
\end{align}
I'm not sure where to go from here (assuming what I've done so far is a reasonable path to the solution).
Can someone help with this ?
 A: For any successor ordinal $\:\alpha<\omega^2$, there is
\begin{align}
(\omega+1)\cdot\alpha&=\underbrace{(\omega +1)+(\omega +1)+\cdots+(\omega +1)}_{\alpha}
\\
&=\underbrace{\omega +(1+\omega)+\cdots+(1+\omega) }_{\alpha}+1
\\
&=\underbrace{\omega +\omega+\cdots+\omega}_{\alpha}+1
\\
&=\omega \cdot \alpha+1
\\
&<\omega^3
\end{align}
Since for any successor ordinal $\:\gamma<\omega^3$, there is a successor ordinal $\:\alpha<\omega^2\:$ that
$$
(\omega+1) \cdot \alpha=\omega \cdot \alpha+\alpha>\gamma\quad\text{and}\quad(\omega+1) \cdot \alpha<\omega^3
$$
So we have
$$
(\omega +1)\cdot \omega^2=\sup\{(\omega+1)\alpha:\alpha<\omega^2\}=\omega^3
$$
A: Following on from where you left off:
For finite $n$, we have
\begin{align*}
(\omega+1)\cdot n &= \underbrace{(\omega+1)+(\omega+1)+ \dots + (\omega+1)}_{n\text{ times}}\\
&= \omega + \underbrace{(1+\omega) + \dots + (1+\omega)}_{n-1\text{ times}}+1\\
&= \omega\cdot n + 1.
\end{align*}
So $(\omega+1)\cdot \omega = \sup_{n\in \omega} (\omega\cdot n + 1) =  \sup_{n\in \omega} (\omega\cdot n) = \omega^2$. (The second equality follows from the observation that $\omega\cdot n \leq \omega\cdot n+1\leq \omega\cdot (n+1)$ for all $n$, so the sequences are mutually cofinal.)
Thus $(\omega+1)\cdot \omega^2 = ((\omega+1)\cdot \omega)\cdot \omega = \omega^2\cdot \omega = \omega^3$.
