algebraic geometry – A question about a nondegenerate smooth curve in $Bbb P^n$ of degree $n$

Suppose that $Xsubset Bbb P^n$ is a nondegenerate smooth projective curve of degree $n$. Let $H$ be a hyperplane in $Bbb P^n$, $D=text{div}(H)$, and $Qsubset |D|$ be the linear subsystem of hyperplane divisors. Then we have the equalities $$ n=dim (Q)leq dim |D|=dim L(D)-1leq n$$
Therefore $Q=|D|$ and $dim L(D)=1+deg (D)$. The latter equality implies that $X$ has genus zero, and the first equality implies that any divisor of degree $n$ is a hyperplane divisor.

This is a paragraph in p.217 of Miranda’s book Algebraic Curves and Riemann Surfaces, and I can’t understand the last sentence.

  1. How do we know that the genus of $X$ is zero? I can only see that by Riemann-Roch $g=dim L(K-D)$.

  2. How do we know that any divisor of degree $n$ is a hyperplane divisor? Is any positive divisor of degree $n$ contained in $|D|$?