Remember that when you multiply order types, you have the left but not the right distributive law. Thus, $(1+\eta)\cdot(\eta+1)=(1+\eta)\cdot\eta+1+\eta$, and $(\eta+1)\cdot(1+\eta)=\eta+1+(\eta+1)\cdot\eta$. For any further simplification, we must see what $(1+\eta)\cdot\eta$ and $(\eta+1)\cdot\eta$ look like.
Suppose that $\langle X,\preceq\rangle$ is a linear order of type $(1+\eta)\cdot\eta$. Then by the definition of $(1+\eta)\cdot\eta$ we can decompose $X$ into pairwise disjoint sets $X_q$ for $q\in\Bbb Q$ in such a way that
- if $p,q\in\Bbb Q$ with $p
, $x\in X_p$, and $y\in X_q$, then $x\prec y$, and
- each $X_q$ is ordered in type $1+\eta$ by $\preceq$ and hence is order-isomorphic to $\Bbb Q\cap[0,1)$ with its usual order.
Note that each $X_q$ is densely ordered by $\preceq$: if $x,y\in X_q$ with $x\prec y$, there is a $z\in X_q$ such that $x\prec z\prec y$.
Clearly $X$ is countably infinite. Now let $x\in X$ be arbitrary; $x\in X_q$ for some $q\in\Bbb Q$. Choose any $y\in X_{q-1}$ and $z\in X_{q+1}$ and note that $y\prec x\prec z$, so that $x$ is neither a first or last element of $\langle X,\preceq\rangle$. Since $x\in X$ was arbitrary, $\langle X,\preceq\rangle$ has no first or last element.
Finally, let $x,y\in X$ with $x\prec y$; there are $p,q\in\Bbb Q$ such that $x\in X_p$ and $y\in X_q$. Clearly $p\le q$. If $p, let $r$ be any rational number between $p$ and $q$; then for each $z\in X_r$ we have $x\prec z\prec y$. If $p=q$, use the earlier observation that $X_p$ is densely ordered to conclude that there is a $z\in X_p$ such that $x\prec z\prec y$. It follows that $\preceq$ densely orders $X$.
Thus, $\langle X,\preceq\rangle$ is a dense linear order without endpoints and is therefore order-isomorphic to $\Bbb Q$; that is, it has order type $\eta$.
We can now conclude that $(1+\eta)\cdot(\eta+1)=(1+\eta)\cdot\eta+1+\eta=\eta+1+\eta$. Now the set $\Bbb Q^-$ of negative rationals has order type $\eta$, as does the set $\Bbb Q^+$ of positive rationals, so $\Bbb Q=\Bbb Q^-\cup\{0\}\cup\Bbb Q^+$ has order type $\eta+1+\eta$, and $\eta+1+\eta=\eta$. (Or you could argue as before that $\eta+1+\eta$ is a countable dense order type without endpoints.) The final conclusion is that $(1+\eta)\cdot(\eta+1)=\eta\;.$
That was pretty detailed, and the other one is quite similar, so I’ll leave it to you.