Is there a more elementary proof of this special case of Riemann-Roch?

I'm looking for an elementary proof of the fact that $\ell(nP) = \dim L(nP) = n$, where $L(nP)$ is the linear (Riemann-Roch) space of certain rational functions associated to the divisor $nP$, where $n > 0$ is an integer and $P$ is a point on a curve of genus 1.

This fact is used to prove that every elliptic curve is isomorphic to a curve given by a Weierstrass equation. See for instance Silverman, The Arithmetic of Elliptic Curves, p. 59, Proposition 3.1.(a).

I know that it's a corollary to the Riemann-Roch theorem (it follows from the corollary that says if $\deg D > 2g - 2$ then $\ell(D) = \deg D + 1 - g$). The proof of the full Riemann-Roch theorem still looks a little intimidating to me, so I'm hoping someone can provide a more elementary proof of this special case, perhaps by making the relevant simplifications in the proof of the Riemann-Roch theorem.

Thanks in advance.

Let $X$ be an elliptic curve over $k$, and $P \in X$. We have to prove that for $D=nP$ we have $$ l(D)=\deg(D)=n $$ If $D$ satisfies the above equation, and $E$ is a divisor such that $E \geq D$, then we have $l(E)=\deg(E)$ (cf. Fulton, Corollary 8.3.1). Hence it is enough to show that $l(P)=1$.

Clearly $l(P) > 0$, since $k \subset L(P)$. On the other hand, $l(P) > 1$ would imply that there exists $x \in k(X)$ such that $x$ has a simple pole at $P$, and no other poles. But this would imply that the map $$ x: X \rightarrow \Bbb{P}^1 $$ is an isomorphism, which is a contradiction. Thus $l(P)=1$.

Here is an elementary solution in a very special case. It assumes that the curve is already embedded in the plane using a Weierstrass-like equation, so you can't use it to later prove that every elliptic curve can be given by a Weierstrass equation, but perhaps it will still be instructive.

We assume the base field is of characteristic $0$ and algebraically closed. Suppose $E$ is the projective curve in $\mathbb{P}^2$ defined by the following equation in $\mathbb{A}^2$, $$y^2 = (x - \lambda_1)(x - \lambda_2)(x - \lambda_3)$$ where $\lambda_1, \lambda_2, \lambda_3$ are non-zero and distinct. It can be shown that $E$ is a smooth curve with a unique point at infinity, namely $P = (0 : 0 : 1)$. As Nils Matthes has indicated, to show that $\ell (n P) = n$ for all $n \ge 1$, it suffices to prove the case $n = 1$.

Suppose, for a contradiction, that $\ell (P) > 1$. We know that for all divisors $D$, $\ell (D) \le \deg D + 1$, so that means $\ell (P) = 2$, and so there is some non-constant rational function $t$ such that $L (P)$ is spanned by $1$ and $t$. Moreover, $t$ has a simple pole at $P$ (and nowhere else), so $t^n$ has a pole of order $n$ at $P$ (and nowhere else). As such, $L (n P)$ is spanned by $1, t, \ldots, t^n$. However, we know that $y$ is in $L (2 P)$ and $x$ is in $L (3 P)$, so that means, for some constants, \begin{align} x & = a_0 + a_1 t + a_2 t^2 \\ y & = b_0 + b_1 t + b_2 t^2 + b_3 t^3 \end{align} and by rescaling $t$ if necessary, we may assume that $a_2 = b_3 = 1$. (Note that $a_2^3 = b_3^2$.) We may also complete the square to assume that $a_1 = 0$. But then substituting into the original equation, we get $$(b_0 + b_1 t + b_2 t^2 + t^3)^2 = (t^2 + a_0 - \lambda_1) (t^2 + a_0 - \lambda_2) (t^2 + a_0 - \lambda_3)$$ which is impossible, since the RHS has six distinct zeros.