In my humble opinion: the following proof is not "very easy" (cit.); but it is the one that I know (via Hilbert polynomials).
Let $I$ be the homogeneous ideal of $\mathbb{K}[x_0,\dots,x_n]$ associated to $X=\{P_0,P_1,\dots,P_m\}$; where $0\leq m\leq kn$, $k\geq2$ and $P_0,P_1,\dots,P_m$ are points in general position; I set $S(X)=\mathbb{K}[x_0,\dots,x_n]_{\displaystyle/I}$.
Without loss of generality, one can assume that $X\cap V_{+}(x_0)=\emptyset$; where $V_{+}(x_0)$ is the hyperplane of equation $x_0=0$. Let
$$
I=(f_1,\dots,f_r),\,J=(f_i(1,x_1,\dots,x_n)\mid i\in\{1,\dots,n\}),
$$
then $S(X)_d\cong S_0(X)=\mathbb{K}[x_1,\dots,x_n]_{\displaystyle/J}$ for $d\gg0$.
Remark. Since $S_0(X)$ is a Noetherian ring of Krull dimension $0$, it is an Artinian ring; in particular, $S(X)$ is a finite-dimensional $\mathbb{K}$-vector space.
Let $h_X$ the Hilbert function of $X$ and let $\chi_I$ the Hilbert polynomial of $I$; because $\dim_{Krull}X=0$ then $\chi_I$ is a constant, by theory:
$$
\exists d_0\in\mathbb{N}_{\geq0}\mid\forall d\in\mathbb{N}_{\geq d_0},\,h_X(d)=\dim_{\mathbb{K}}S(X)_d=\chi_I;
$$
defined
$$
\mathbb{K}[x_1,\dots,x_n]_{\leq d}=\{f\in \mathbb{K}[x_1,\dots,x_n]\mid\deg f\leq d\},\,J_{\leq d}=\{f\in J\mid\deg f\leq d\},
$$
let $B_d$ a base of $J_{\leq d}$ as $\mathbb{K}$-vector space: by remark this is a finite set, then
$$
^hB_d=\left\{x_0^df\left(\frac{x_1}{x_0},\dots,\frac{x_n}{x_0}\right)\in I_{\leq d}\mid f\in B_d\right\}
$$
is a base of $I_{\leq d}$ as $\mathbb{K}$-vector space; in particular:
$$
\dim_{\mathbb{K}}J_{\leq d}=\dim_{\mathbb{K}}I_{\leq d};
$$
from all this:
$$
\forall d\geq d_0,\,\chi_I=h_X(d)=\dim_{\mathbb{K}}\mathbb{K}[x_1,\dots,x_n]_{\leq d\displaystyle/J_{\leq d}}.
$$
Considering the canonical epimorphism
$$
\pi:\mathbb{K}[x_1,\dots,x_n]\twoheadrightarrow\mathbb{K}[x_1,\dots,x_n]_{\displaystyle/J},
$$
it induces the inclusion
$$
i:\mathbb{K}[x_1,\dots,x_n]_{\leq d\displaystyle/J_{\leq d}}\hookrightarrow\mathbb{K}[x_1,\dots,x_n]_{\displaystyle/J};
$$
let $E$ be a base of $\mathbb{K}[x_1,\dots,x_n]_{\displaystyle/J}$, then $E_{\leq d}=\{f\in E\mid\deg f\leq d\}$ is a system of generators for $\mathbb{K}[x_1,\dots,x_n]_{\leq d\displaystyle/J_{\leq d}}$, that is one can define an epimorphism
$$
\epsilon:\mathbb{K}[x_1,\dots,x_n]_{\displaystyle/J}\twoheadrightarrow\mathbb{K}[x_1,\dots,x_n]_{\leq d\displaystyle/J_{\leq d}}
$$
where $\ker\epsilon=\langle f\in E\mid\deg f>d\rangle$.
Because these are finite-dimensional $\mathbb{K}$-vector spaces, they are isomorphic; and moreover:
$$
\forall d\geq d_0,\,\chi_I=h_X(d)=\dim_{\mathbb{K}}S(X),
$$
in other words, $I$ can be generated by homogeneous polynomials of degree at most $d_0$.
Important remark. Until this point: I have not used the fact that $X$ is a (finite) set of points in general position!
Let $X=Y_1\cup\dots\cup Y_k$ a partion in sets with at most $n$ points, then any $Y_i$ is contained in an hyperplane $H_i$; let $f_i$ be the linear polynomial vanishing on $H_i$, since $F=f_1\cdot\dots\cdot f_k\in I$ and $\deg F=k$ one has $k\geq d_0$; in other words, $I$ can be generated by polynomials of degree at most $k$.
