According to Wolfram,
If $\beta$ is a limit ordinal, then $\alpha*\beta$ is the least ordinal greater than any ordinal in the set $\{\,\alpha*\gamma:\gamma<\beta\,\}$ (Suppes 1972, p. 212).
So with $\beta=\omega$ and $\alpha=0$, we see that $\{\,0*n:n<\omega\,\}=\{0\}$ and would infer that $0* \omega=1$. This seems to be a weird claim (and backed up by literature?), given that the definition via lexical ordering of cartesian products clearly results in $\alpha*0=0*\alpha=0$ for all $\alpha$.
Is that just a typo or are there competing definitions of ordinal multiplication that differ in this respect common?
They mean to define $\alpha \ast \beta$ for every fixed $\alpha$ by recursion on $\beta$.
So we need a starting clause $\alpha \ast 0 = 0$.
A successor clause: $\alpha \ast (\beta+1) = \alpha \ast \beta + \beta$, also no problem.
A limit clause (but in words it was messed up):
For a limit ordinal $\beta$, such that $\alpha \ast \gamma$ is defined for all $\gamma < \beta$:
$$\alpha \ast \beta = \sup \{\alpha \ast \gamma: \gamma < \beta\}$$
and all is well in the world. It's now easy to show that $0 \times \omega = 0$ as it ought to be.
It's an interesting exercise now to show that this coincides with the more direct way of defining it by taking the order type of $\alpha \times \beta$ in reverse lexicographic order ($(x,y) <(u,v)$ iff $ y < v$, or $y=v$ and $x < u$).