Proposition: Let $P_0,\ldots,P_{n+1}$ be $n+2$ points in $\mathbb{P}^n$ such that every $n+1$ are in general position. There exists a coordinate system in $\mathbb{P}^n$ such that $P_0=(1:0:\cdots:0),\ldots,P_n=(0:0:\cdots:1),P_{n+1}=(1:\cdots:1)$.
I found a proof but I have some doubts. Let me write it.
Proof: Fix a homogeneous coordinate system in $\mathbb{P}^n$. Denote $P_k=(a_0^k:\cdots:a_n^k)$ for every $k=0,\ldots,n$. Denote $u_k=(a_0^k,\cdots,a_n^k) \in \mathbb{K}^{n+1}$. As $P_0,\ldots,P_n$ are in general position, $\{u_0,\ldots,u_n\}$ is a basis of $\mathbb{K}^{n+1}$. Then $u_{n+1}=\sum_{k=0}^n \alpha_k u_k$ with $\alpha_k \neq 0$ for all $k$. Hence in the basis $\{\alpha_0 u_0,\ldots,\alpha_n u_n\}$ of $\mathbb{K}^{n+1}$, the homogeneous coordinates of each $P_k$ is what is given in the proposition.
My doubts:
Why is $\alpha_k \neq 0$ for all $k$? Is anything about the general position? [Closed]
More or less I understand why $\{\alpha_0 u_0,\ldots,\alpha_n u_n\}$ is the required basis. Formally we have $$(1,0,\ldots,0)=1 \cdot \alpha_0 u_0 = \alpha_0(a_0^0,\ldots,a_n^0)$$ $$\vdots$$ $$(0,\ldots,0,1)=1 \cdot \alpha_n u_n = \alpha_n(a_0^n,\ldots,a_n^n)$$ $$(1,\ldots,1)=1 \cdot \alpha_0 u_0 + \cdots + 1 \cdot \alpha_n u_n = u_{n+1} = (a_0^{n+1},\ldots,a_n^{n+1})$$ Is this enough to show the last sentence of the proof?
Thanks in advance.