Constant morphism between projective varieties i'm currently working my way through Algebraic Geometry and stuck at Exercise 7.9 b)

Let $f: \mathbb{P}^n \to \mathbb{P}^m$ be a morphism Prove:
  If $ n > m $ then $f$ must be constant

I found this old thread here and would like to proceed by showing that for an affine subset $\mathbb{A}^n \subset \mathbb{P}^n$ we know that the morphism $f$ restricted on the subset has the form $f(x) = (f_1(x),...,f_m(x))$ for homogeneous $f_i$. So i would like to show what was mentioned in the first commentar that the $V(f_1,...,f_m)\setminus \{ 0 \}$ is not empty and hopefully inside my affine subset.
Would this be a possible solution for this exercise and how can i do it?
 A: You are right, you can argue as in the comment you pointed at. In the following, I will try to assume the content of the notes you are following.
A morphism $f \colon \mathbb{P}^n \rightarrow \mathbb{P}^m$ is given by $m+1$ homogeneous polynomials of the same degree. This is what the comment you quoted says in the special case $m=1$, and is close to the content of Lemma 7.5 in your notes.
By Exercise 6.32, the intersection of $n$ hypersurfaces in $\mathbb{P}^n$ is not empty (provided that $n \geq 2$, which we may assume by the setup of our problem). If we assume $m+1 \leq n$, it follows that the polynomials giving the map to $\mathbb{P}^m$ all vanish at some point, and so the map in not everywhere defined.
It is important to work with $\mathbb{P}^n$ and not an affine chart $\mathbb{A}^n$ to argue that these $m$ polynomials need to have common zeroes. Indeed, this is not the case for a general choice of affine chart. An example for this is the projection away from a point $P$ of $\mathbb{P}^n$ to $\mathbb{P}^{n-1}$. This map is defined everywhere except at $P$. Thus, if we take an affine chart not containing $P$, there the morphism will be described by some polynomials withouth a common zero.
Once you learn the language of line bundles, you can give another perspective to the problem. If we have a morphism $f \colon \mathbb{P}^n \rightarrow \mathbb{P}^m$, we can pull back the line bundle $\mathcal{O}_{\mathbb{P}^m}(1)$ by $f^*$. This gives a line bundle on $\mathbb{P}^n$. On $\mathbb{P}^n$, all the line bundles are $\mathcal{O}_{\mathbb{P}^n}(d)$ for some $d$. Since $\mathcal{O}_{\mathbb{P}^m}(1)$ has more than one global section, it needs to pull back to $\mathcal{O}_{\mathbb{P}^n}(d)$ for $d \geq 1$, giving an ample line bundle. On the other hand, if $m < n$, no line bundle on $\mathbb{P}^m$ can pull back to an ample line bundle for dimension reasons.
