Recall that $\alpha\beta$ is the order type of $\beta$ copies of $\alpha$, one after the other. If $\beta<\gamma$, then obviously ($\gamma$ copies of $\alpha$) is longer than only ($\beta$ copies). This shows that the function $\beta\mapsto \alpha\beta$ is strictly increasing.
The argument showing that the map is continuous is similar: If $\gamma>\alpha\delta$ for all $\delta<\beta$, then $\gamma\ge$ ($\beta$ copies of $\alpha$). [Or you can think of it this way: If $\gamma<\alpha\beta$, then $\gamma$ lies inside (or "just to the left of") one of the $\beta$ intervals of type $\alpha$ that make up $\alpha\beta$, say the $\delta$-th one. Then $\gamma<\alpha\delta$. This shows that $\alpha\beta$ is the sup of the $\alpha\delta$ for $\delta<\beta$.]
Of course, details may vary if the definition of ordinal multiplication you are using is different. (There are several equivalent possibilities.)
Let me expand a bit on why we argued as we did, since this is an issue that comes up with some frequency: The topology here is the order topology on ordinals. Note that $0$ and all successor ordinals are isolated: $\{\alpha+1\}=(\alpha,\alpha+2)$ and $\{0\}=(-\infty,1)$. On the other hand, any open neighborhood of $\alpha$ limit contains one of the form $(\beta,\alpha+1)$ for some $\beta<\alpha$.
To show that a function $f$ is continuous, we need to show that the preimage of any open set is open. It follows that it suffices to prove that if $f(\lambda)$ is limit, then for any $\beta there is a neighborhood $U$ of $\lambda$ contained in $(\beta,f(\lambda)+1)$. Of course, this is trivial if $\lambda$ itself is not a limit ordinal. On the other hand, if $\lambda$ is limit, we need to check that $\beta for $\gamma<\lambda$ sufficiently large.
Suppose now, as in the question, that $f$ is strictly increasing. Then the condition above reduces to check, for $\lambda$ limit, that $f(\lambda)=\sup\{f(\gamma)\mid\gamma<\lambda\}$.