Why does each subfield have to contain multiples of 1? I have seen a Theorem:

every field contains exactly one prime subfield $K_0$.

and next:

$K_0$ has to contain 1 and all its multiples: $n \cdot 1 = 1 + \ldots + 1$

Is this because it has to contain 0 and 1 by definition, and then it must also holds that $1 + 1$ belongs to K, and $(1 + 1) + 1$ too because of two-arguments $+$ operation of its abelian group?
so we can prove it by induction?

If $A$ is a group and $a\in A$, then there is a unique group homomorphism $f\colon \mathbb Z\to A$ with $f(1)=a$.
We write $a^n$  or - if $A$ is written additively - $n\cdot a$ for the element $f(n)\in A$ with $f$ defined like this,
Apply this to the additive group of $K$ and $a=1$.
