Can you tell me why doing scalar multiplication of a point on a Elliptic curve over a finite field gets to a point at infinity? I am reading Programming Bitcoin. The author said:


  Another property of scalar multiplication is that at a certain multiple, we get to the point at infinity (remember, the point at infinity is the additive identity or $0$). If we imagine a point $G$ and scalar-multiply until we get the point at infinity.


He doesn't explain why. So I don't understand why. I would like you to give me a plain explanation, without a serious mathematical proof, if that could be possible.
The points on the Elliptic Curves $E$ over a finite field $K$ are forming a commutative additive group ( Abelian group) with the point addition as the group operation. The group needs an identity element and usually represented as $\mathcal{O}$. In most curves, this is the point at infinity and also notated as $P_\infty$. In the Edwards curves, the identity is an affine point $(0,1)$.
The scalar multiplication $[k]P$ this actually means adding the $P$ itself $k$-times. More formally;
let $k \in \mathbb{N}\backslash\{ 0\}$
\begin{align}
[k]:&amp; E \to E\\
&amp;P\mapsto [k]P=\underbrace{P+P+\cdots+P}_{\text{$k$ times}}.\end{align}
and by being identity $[0]P = \mathcal{O}$, and $[k]P=[-k](-P)$ for $k&lt;0$.
Since $\#E(K)$, the number of points, on Elliptic curves over a finite field is finite if you add a point $P$ itself many times eventually you will get the identity $\mathcal{O}$.
$$\underbrace{P+P+\cdots+P}_{\text{$t$ times}} = [t]P= \mathcal{O}$$
The smallest $t$ will be the order of the subgroup generated by the $P$. For security, we want this order huge since the attacks. If we consider the  expected attack cost for Pollard's $\rho$ is $O(\sqrt{\ell})$, i.e. we need to double the point size.
Bitcoin uses Secp256k1 which has characteristic $p$ and it is defined over the prime field $\mathbb{Z}_p$ with the curve equation $y^2=x^3+7$.
The order $n$ of the base point
G = 02 79BE667E F9DCBBAC 55A06295 CE870B07 029BFCDB 2DCE28D9 59F2815B 16F81798
is
n = FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFE BAAEDCE6 AF48A03B BFD25E8C D0364141
Note 1: a point $P$ may not generate the whole group but it generates a cyclic subgroup.
Note 2: As pointed by SqueamishOssifrage, The Smart showed that if the order of the curve and order of the base field ($K$) are same then the discrete logarithm on this curves runs in linear time.