Point-at-infinity in the scalar multiplication Is it possible to hit point-at-infinity while performing scalar multiplication in Weierstrass curves?

According to this reply this is only possible when the scalar k is bigger then the order of the curve n. Is it so? If yes why?
Yes, it is possible.

The smallest scalar $k$ for which $[k]P = \infty$ is called the order of $P$.

For every non-zero point $P$ on the curve there is such a $k$.

This is simply due to the fact that there is only a finite amount of points so if you keep increasing $k$ at some point you must get a repetition $[k]P = [k-i]P$ for some $i$. Now $[k]P - [k-i]P = \infty \iff [i]P = \infty$.

No, $k$ does not have to be bigger than the group order $n$.

In fact for any prime $p$ dividing $n$ there is a point $Q$ such that $[p]Q = \infty$. This is Cauchy's theorem.

On the other hand Lagrange's theorem tells us that the order of every point divides the group order: $[k]P = \infty \Rightarrow k|n$. Therefore, if the group order $n$ is picked to be prime it follows that the order of every non-zero element is $n$.