Why is $\phi:\mathbb{P}^n\rightarrow \mathbb{P}^m$ constant if dim $\phi(\mathbb{P}^n)<n$?

Let $\phi:\mathbb{P}^n\rightarrow \mathbb{P}^m$, $n\leq m$. I want to demonstrate that if dim $\phi(\mathbb{P}^n)<n$ then $\phi(\mathbb{P}^n)=pt$ (ex. 7.3(a), ch.II from Hartshorne).

It’s well known that $Pic(\mathbb{P}^n)\simeq\mathbb{Z}$, with generator $O_{\mathbb{P}^n}(1)$.

First question : it is right that if I show that $\phi^*O_{\mathbb{P}^m}(1)$ is generated by less than n+1 global sections, then must be $\phi^*O_{\mathbb{P}^m}(1)\simeq O_{\mathbb{P}^n}$ and so $\phi(\mathbb{P}^n)=pt$ ?

Second question: if so, let $s_0,…,s_m$ be global sections of $O_{\mathbb{P}^m}(1)$. Can I say that, because of the dimension hypotesis, $\phi^*O_{\mathbb{P}^m}(1)$ can be generated by $\phi^*s_{i_0},…,\phi^*s_{i_r}$ with $r<n$ ? Why?

Solutions Collecting From Web of "Why is $\phi:\mathbb{P}^n\rightarrow \mathbb{P}^m$ constant if dim $\phi(\mathbb{P}^n)<n$?"

The answer to your first question is yes. The line bundle $\mathcal{O}(d)$, with $d > 0$, is generated by the degree $d$ monomials of which there are $\binom{n+d}{d}$. So the number of generators you need grows with $d$.

Your second question is a bit too naive. In general, if you have a morphism $X\to \mathbb{P}^{m}$, there is no link between the dimension of the image and the number of generators of the induced line bundle.

However, by the above we know that there cannot be a nonconstant morphism $\mathbb{P}^{n}\to \mathbb{P}^{m}$ with $m<n$.
Now if we have $n\leq m$ and $dim(\phi(\mathbb{P}^{n}))<n$ we can repeatedly project until we get a map $\tilde{\phi}:\mathbb{P}^{n}\to \mathbb{P}^{n-1}$.
Since projecting away from a point is a finite map, we have $dim(\phi(\mathbb{P}^{n}))=dim(\tilde{\phi}(\mathbb{P}^{n}))$, and we’re done.