It is well known that meromorphic functions in $\mathbb P^n_\mathbb C$ are of the form $p/q$ where $p$, $q$ are homogeneous polynomials in $\mathbb C[x_0 , \ldots , x_n]$ of the same degree, and $q \neq 0$. For example here it is shown using Chow's theorem that subvarieties of $\mathbb P^n_\mathbb C$ are algebraic.
However, reading the book Complex geometry: An introduction by D. Huybrechts, Problem 2.1.3 asks to find the algebraic dimension $a(\mathbb P^n)$ of $\mathbb P^n$, which is the transcendence degree of the field of meromorphic functions. At this point in the book, only the local theory of holomorphic functions and the definitions of complex manifolds have been carried put, so I am wondering if there is an elementary proof of this (ie, that the $a(\mathbb P^n)=n$), since Chow's theorem is very deep.
By elementary I mean, using basically the local Theorey of holomorphic functions in $\mathbb C^n$ ie Hartogs' Theorem, Weierstrass prparation and its corollaries, Riemann extension theorem and the Nullstellensatz.
It is clear that in order to get $a(\mathbb P^n)=n$ it is enough to show that the meromorphic functions are rational, as explained in the comments. I would imagine this can be done using line bundles, the fact that any line bundle in $\mathbb P^n$ is $\mathcal O(d)$ and that sections of that line bundle are polynomials of degree $d$. This approach uses Hartogs' Theorem. However, it uses material which is much more advanced in the book but it hints to me that one can try to use Hartogs' Theorem to give the elementary proof I am looking for.
EDIT: An answer should by an elementary proof of one of the two facts: $\mathbb P^n$ has algebraic dimension $n$ or all meromorphic functions in $\mathbb P^n$ are quotients of polynomials of the same degree.