The degree of smooth projective curve included in $\mathbb P^n$ which is nondegenerate is more or equal the dimension of the projective space

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I had a theorem during lecture, with proof which I don't understand. Theorem says:

$ X \subset \mathbb{P}^n(\mathbb{C})$ smooth projective curve which is nondegenerate (not contained in a hyperplane $\mathbb{P}^{n-1}(\mathbb{C})$) then $\deg(X) \geqslant n $

where $\deg(X)$ for a smooth projective curve $X$ is equal $\deg(i^*H)$ where $i: X \rightarrow \mathbb{P}^n(\mathbb{C})$ is inclusion map and $H$ is a hyperplane divisor of $X$

Proof:

We have proven that $D \in \textrm{Div}(X) \Rightarrow \dim(L(D)) \leqslant deg(D) +1 \Rightarrow dim(|D|) \leqslant \deg(D) $ so let's take the inclusion $i: X \rightarrow \mathbb{P}^n(\mathbb{C})$ and $D=i^*H$ . In general $n \leqslant \dim(|D|) \Rightarrow n \leqslant \deg(D)$.

My question is:

why do we know that "In general $n \leqslant \dim(|D|)$ " ?