Please could you look at the demonstration I propose below for this implication and tell me if I am wrong.
What other way or idea is there to demonstrate the implication?
Proof:
The implication is equivalent to showing that:
if $\alpha$ is limit or $\beta$ is limit $\Rightarrow$ $\alpha \beta$ is limit
Proof by reductio ad absurdum, be $\alpha \beta$ successor, them $\alpha \beta=\lambda+1$.
Case $\alpha$ limit:
$\alpha=\sup \{\gamma: \gamma<\alpha\}$ them for all $\gamma<\alpha$ so that $\gamma+1<\alpha$. In particular let $\alpha \beta=\lambda+1$ $=\gamma+1$, them $\alpha \beta<\alpha$ and it's false; because, let $1<\beta$ them $\alpha<\alpha \beta$ when $0<\alpha$. If $\beta=0$ or $\beta=1$ then $\alpha \beta$ is limit. and if $\alpha$=0 then $\alpha \beta<0$, it's false