I'm trying to see that $A_k(\mathbb{P}^n)=\mathbb{Z}$ for all $k$. I am trying to do this with induction on $n$, by applying the excision sequence $$A_k(Y) \xrightarrow{i_*} A_k(X) \xrightarrow{j^*} A_k(U) \to 0$$ with $U=\mathbb{P}^n\setminus\mathbb{P}^{n-1} \cong \mathbb{A}^n$, and using the fact that the only nontrivial Chow group of $\mathbb{A}^n$ is $A_n(\mathbb{A}^n) \cong \mathbb{Z}$.
Now, I understand the case $n=1$, and obviously $A_n(\mathbb{P}^n) \cong \mathbb{Z}$. For $k<n$, we get that $A_k(\mathbb{P}^{n-1}) \xrightarrow{i_*} A_k(\mathbb{P}^n)$ is surjective, so it remains to show that $i_*$ is injective. By induction, $A_k(\mathbb{P}^{n-1})$ is generated by the class of a $k$-dimensional subscheme $Y \subseteq \mathbb{P}^{n-1}$. Assume that there are finitely many $(k+1)$-dimensional varieties $V_1,\ldots,V_t$ of $\mathbb{P}^n$, and $r_i \in K(V_i)^*$ such that $m[Y]=\sum_{i=1}^t [\mathrm{div}(r_i)]$. Why does this imply that $m=0$?