Multiplication by n on E(K) is surjective

What's the easiest way to see this?

I can imagine a proof for $n=2^k$ since for some $P \in E(K)$ you can just move a line intersecting P round the curve till it's tangent, then that point, say $Q \in E(K)$ would be such that $2Q=P$ and induction would get the rest.

Struggling to see the proof for n not a power of 2 however.

• This is not generally true, e.g., if $K$ is the rational numbers. It would help readers if you defined all your notation (what's $E$?), made the content of the question independent of the subject line of the question, and included any necessary conditions that you ignored, thereby ruling out my counterexample. – KCd Feb 17 '15 at 16:59
• This is stille true for abelian varieties (with $K$ algebraically closed): math.stackexchange.com/questions/1784807 – Watson Nov 17 '18 at 14:05

The claim is true if $K$ is algebraically closed, and it follows from a more general fact about algebraic curves: every map $\phi\colon X\to Y$ of algebraic curves over an algebraically closed field is either constant or surjective. So you just need to show that multiplication by $n$ is not constant, which is a very easy exercise.