Index of speciality of general line bundles I found the following statement

Let $X$ be a smooth curve of genus $g$ over an algebraically closed field. If $\mathscr{L}$ is a general line bundle of degree $d\geq g-1$ the $\mathscr{L}$ is non-special, i.e. $H^1(\mathscr {L})=0$.

I don't figure how to prove it with simple arguments (to be clear, not so deeper than Riemann-Roch); Eisenbud suggest a proof that involves the Picard variety of $X$, an approach that I don't like because it seems a bit involved.
Hope to have some hints to find a simpler way to prove it.
 A: I think one can reason like this, without using the Picard variety. First we need to understand what "general" means. We can work with divisors, and say that a line bundle $\mathscr{L}$ is general if $\mathscr{L}=\mathcal{O}(p_1+\dots+p_r-q_1-\dots-q_s)$, with $p_i,q_j$ general points in $X$ 
Now we can prove the statement.

Remark: The statement is equivalent to proving that for a general line bundle $\mathscr{L}$ of degree $d\geq g-1$ we have $h^0(\mathscr{L})=d-(g-1)$.

Proof of Remark: this is simply Riemann-Roch.

Lemma Let $\mathscr{M}$ be a line bundle of such that $h^0(\mathscr{M})\geq r$. Then for general points $p_1,\dots,p_r$ we have $h^0(\mathscr{M}(-p_1-\dots-p_r))=h^0(\mathscr{M})-r$.

Proof of Lemma: by induction on $r$. If $r=0$ there is nothing to prove. Then we suppose that $r>0$ and that the statement is true for $r-1$. Since $H^0(\mathscr{M})>0$, a general point $p_r\in C$ does not belong to the base locus of $\mathscr{M}$, so that $h^0(\mathscr{M}(-p_r))=h^0(\mathscr{M})-1 \geq r-1$. Then by inductive hypothesis we know that for general points $p_1,\dots,p_{r-1}\in X$ we have $h^0(\mathscr{M}(-p_1-\dots-p_r))=h^0(\mathscr{M}(-p_r))-(r-1)=h^0(\mathscr{M})-r$.
Proof of Statement: First we observe that if $\mathscr{L}$ is any line bundle of degree $d\geq 2g-1$, then $h^1(\mathscr{L})=h^0(\omega_X\otimes \mathscr{L}^{-1})=0$ since $\deg(\omega_X\otimes \mathscr{L}^{-1})<0$. Then the statement holds and the remark shows that $h^0(\mathscr{L})=d-(g-1)$. Now, fix $g-1 \leq d \leq 2g-2$ and let $\mathscr{M}$ be a general line bundle of degree $2g-1$. Then choose general points $p_1,\dots,p_{2g-1-d}$ and set $\mathscr{L}=\mathscr{M}(-p_1-\dots-p_{2g-1-d})$. We see that  $h^0(\mathscr{M})=2g-1 - (g-1)=g$. In particular $h^0(\mathscr{M}) \geq 2g-1-d$ so that we can use the Lemma and obtain
$$ h^0(\mathscr{L}) = h^0(\mathscr{M}(-p_1-\dots-p_{2g-1-d})) = h^0(\mathscr{M}) - (2g-1-d) = g - 2g +1 +d =g-(d-1)  $$
Then we are done using the remark.
