The following argument is, I believe, based on the premise that an isogeny (or a morphism of curves that is a group homomorphism) doesn't ramify:
Considering the multiplication by $n$ map $[n]$ on a elliptic curve $E/K$, where char$(K)\not|\,\,n$ so that $[n]$ is separable:
$|[n]^{−1}(Q)| ≤ deg[n] = n^2$ with equality for all but finitely many $Q ∈ E.$ But $[n]$ is a group homomorphism, so $|E[n]| = n^2$
$E[n]$ is by definition $|[n]^{−1}(\mathcal O_E)|$ so this is at least saying that $\mathcal O_E$ is not one of those points where the map ramifies (but I believe this implies that no other point ramifies). Why is this so?
In positive characteristics, isogenies may ramify. In general, see Silverman's "The Arithmetic of Elliptic Curves", Theorem 4.10 in Chapter III, where he shows that if $\phi:E_1\to E_2$ is a non-constant isogeny then
(1) For every $Q\in E_2$, we have $\# \phi^{-1}(Q)$ equals the separable degree of $\phi$.
(2) For every $P\in E_1$, the ramification index of $\phi$ at $P$ equals the inseparable degree of $\phi$.
(3) If $\phi$ is separable, then $\phi$ is unramified and $\# \ker(\phi)$ is equal to the (separable) degree of $\phi$.
The proof of (1) uses the fact that if you have a non-constant map of curves, then $\# \phi^{-1}(Q)$ equals the separable degree for all but finitely many $Q$, together with the fact that $\phi$ is a group homomorphism, to show that in fact $\# \phi^{-1}(Q)=\deg_s(\phi)$ holds for all $Q$.