Consider $\Bbb P^{n+1}$ with coordinates $t,u_0,u_1,\cdots,u_n$ and $\Bbb P^n$ with coordinates $u_0,\cdots,u_n$. There is an holomorphic fibre bundle: $$\pi:\Bbb P^{n+1}\setminus{[1:0:\cdots:0]}\to\Bbb P^n\\\,[t:u_0:u_1:\cdots:u_n]\mapsto[u_0:u_1:\cdots:u_n]$$ which is verifiably a model for $\mathcal{O}(1)$. The best way to do this is to check the transition data; using the canonical charts $U_i=\{u_i\neq0\}$, $\pi$ is locally the projection $\Bbb C^{n+1}\to\Bbb C^n$ which forgets the first coordinate, where the axes are: $t/u_i,u_0/u_i,\cdots,\hat{u_i},\cdots,u_n/u_i$ and $u_0/u_i,\cdots,\hat{u_i},\cdots,u_n/u_i$. The transition functions are the correct ones, namely $\psi_{ij}=u_j/u_i$ where $f_i=\psi_{ij}\cdot f_j$ must hold for holomorphic local sections.
Let $\sigma:\Bbb P^n\to\Bbb P^{n+1}\setminus[1:\mathbf{0}]$ be an holomorphic section of $\pi$. If $\sigma$ is associated to a linear functional $\varphi$, then $\sigma$ would be the map $[u_0:u_1:\cdots:u_n]\mapsto[\varphi(u_0,\cdots,u_n):u_0:\cdots:u_n]$. As it is, on each chart $U_i$ we can consider the maps $(z_1,\cdots,z_n)\mapsto[z_1:\cdots:1:\cdots:z_n]\overset{\sigma}{\mapsto}[f_i(z):z_1:\cdots:1:\cdots:z_n]$ where the "$1$" goes in the $i$th slot. This well-defines the holomorphic function $f_i:\Bbb C^n\to\Bbb C$.
Thinking about the equality $[z_1:\cdots:1:\cdots:z_n]=[1:z_2/z_1:\cdots:1/z_1:\cdots]$, we have: $$f_i(z)/z_1=f_0(z_2/z_1,\cdots,1/z_1,\cdots,z_n/z_1)$$for $z_1\neq0$ and $i\neq 0$, etc. so that a kind of homogeneity is enforced. Again, this equation essentially contains a proof that $\Bbb P^{n+1}\setminus\{\text{pt}\}\to\Bbb P^n$ models $\mathcal{O}(1)$ as we can infer the transition functions are what they should be.
The base case where $n=0$ is absolutely trivial. By induction, $\sigma$'s restriction to the hyperplane $u_0=0$ comes from a linear functional $\psi$ on $\Bbb C^n$. Let $\tau$ be the section associated to $\psi$ viewed as a linear functional on $\Bbb C^{n+1}$, forgetting the first coordinate, and $g_i$ its associated holomorphic functions as above. $h=f_0-g_0$ is an holomorphic function on $\Bbb C^n$ with $h(z)/\|z\|\to0$ at infinity (e.g. $[t:1:z_1:\cdots]=[t/z_1:1/z_1:1:\cdots]\to[0:0:1:\cdots]$ as $z_1\to\infty$ since $[0:1:\cdots]$ is in the domain where $t=f-g$ vanishes as $\sigma,\tau$ agree) therefore $h$ is a polynomial of total degree lesser or equal to $1$ - in fact, constant - and $g_0$ in fact being a linear form (yes, really, but that relies on $\psi$ being extended by zero!) entails that $f_0$ is also a polynomial of total degree lesser or equal to $1$. The same would be true for any of the $f_j$, of course.
Let $f_0(z)=\alpha+\sum_i c_iz_i$. It's not too important, but here $\alpha=h$ would be the constant and $(c_1,c_2,\cdots,c_n)$ would represent $\psi$.
Then the equation: $$f_i(z)=z_1\cdot f_0(z_2/z_1,\cdots,1/z_1,\cdots,z_n/z_1)$$finds: $$f_i(z)=\alpha z_1+\sum_{j<i+1}c_j z_j+c_{i+1}+\sum_{j>i+1}c_j z_{j-1}$$and it becomes clear that $\sigma$ globally arises from $\varphi$ the linear form with vector $(\alpha,c_1,c_2,\cdots,c_n)$.
