Promotion 4 Let $n$ be a positive integer, consider
$$
\begin{aligned}
S= & \left\{\left(x_{1}, x_{2}, \cdots, x_{k}\right) \mid x_{1}, x_{2}, \cdots, x_{k} \in\{0,\right. \\
& \left.1, \cdots, n\}, x_{1}+x_{2}+\cdots+x_{k}>0\right\}
\end{aligned}
$$

is a set of $(n+1)^{k}-1$ points in $k$-dimensional space. Try to find the minimum number of hyperplanes whose union contains $S$ but does not contain $(0,0, \cdots, 0)$.
Definition 3 For the function $f\left(x_{1}, x_{2}, \cdots, x_{k}\right)$ over $\mathbf{R}[x]$, the first-order difference is
$$
\begin{array}{l}
\Delta_{x_{1}} f=f\left(x_{1}+1, x_{2}, \cdots, x_{k}\right)-f\left(x_{1}, x_{2}, \cdots, x_{k}\right), \\
\Delta_{x_{2}} f=f\left(x_{1}, x_{2}+1, \cdots, x_{k}\right)-f\left(x_{1}, x_{2}, \cdots, x_{k}\right), \\
\cdots \cdots \\
\Delta_{x_{k}} f=f\left(x_{1}, x_{2}, \cdots, x_{k}+1\right)-f\left(x_{1}, x_{2}, \cdots, x_{k}\right),
\end{array}
$$

where $\Delta \in\left\{\Delta_{x_{1}}, \Delta_{x_{2}}, \cdots, \Delta_{x_{k}}\right\}$ is called the difference operator.
Theorem 2 Newton's formula:
$$
\Delta^{n}=\sum_{i=0}^{n}(-1)^{n-i} \mathrm{C}_{n}^{i} E^{i}(n=0,1, \cdots) \text {. }
$$

Theorem 3 Let $f(x)$ be a polynomial of degree $m$ with leading coefficient $a_{n}$. If the positive integer $n>m$, then $\Delta^{n} f(x) \equiv 0$, and
$$
\sum_{i=0}^{n}(-1)^{n-i} \mathrm{C}_{n}^{i} f(i)=0 \text {. }
$$

Returning to Generalization 4.
Proof Assume there exist $m$ hyperplanes
$$
\begin{array}{l}
a_{1 i} x_{1}+a_{2 i} x_{2}+\cdots+a_{k i} x_{k}-a_{k+1, i}=0(1 \leqslant i \leqslant m), \\
a_{k+1, i} \neq 0
\end{array}
$$

satisfying the conditions.
Construct the polynomial of degree $m$
$$
\begin{array}{l}
f\left(x_{1}, x_{2}, \cdots, x_{k}\right) \\
=\prod_{i=1}^{m}\left(a_{1 i} x_{1}+a_{2 i} x_{2}+\cdots+a_{k i} x_{k}-a_{k+1, i}\right) .
\end{array}
$$

By the problem statement, for any $\left(x_{1}, x_{2}, \cdots, x_{k}\right) \in S$, we have
$$
f\left(x_{1}, x_{2}, \cdots, x_{k}\right) \equiv 0,
$$

but $f(0,0, \cdots, 0) \neq 0$.
If $m<k n$, by Theorem 3 we get
$$
\Delta_{x_{1}}^{n} \Delta_{x_{2}}^{n} \cdots \Delta_{x_{k}}^{n} f \equiv 0 \text {. }
$$

Also, by Theorem 2 we have
$$
\Delta^{n} f(x)=\sum_{i=0}^{n}(-1)^{n-i} \mathrm{C}_{n}^{i} f(x+i) \text {. }
$$

Generalizing to $k$ dimensions, we get
$$
\begin{array}{l}
\Delta_{x_{1}}^{n} \Delta_{x_{2}}^{n} \cdots \Delta_{x_{k}}^{n} f\left(x_{1}, x_{2}, \cdots, x_{k}\right) \\
=\sum_{\left(j_{1}, j_{2}, \cdots, j_{k}\right.} \sum_{\in S \cup(0,0, \cdots, 0)!}(-1)^{k_{n}-j_{1}-j_{2}-\cdots-j_{k}} . \\
\mathrm{C}_{n}^{j_{2}} \mathrm{C}_{n}^{j_{2}} \cdots \mathrm{C}_{n}^{j_{k}} f\left(x_{1}+j_{1}, x_{2}+j_{2}, \cdots, x_{k}+j_{k}\right) .
\end{array}
$$

Let $x_{1}=x_{2}=\cdots=x_{k}=0$, we get
$$
\begin{array}{l} 
f(0,0, \cdots, 0) \\
= \sum_{\left(j_{1}, j_{2}, \cdots, j_{k}\right) \in s}(-1)^{j_{1}+j_{2}+\cdots+j_{k}+1} . \\
\mathrm{C}_{n}^{j_{1}} \mathrm{C}_{n}^{j_{2}} \cdots \mathrm{C}_{n}^{j_{k}} f\left(j_{1}, j_{2}, \cdots, j_{k}\right) \\
= 0,
\end{array}
$$

which contradicts the problem statement.
Therefore, $m \geqslant k n$.
Next, construct $k n$ hyperplanes that include all points in $S$.

Notice that each point in $S$ has at least one coordinate equal to $1,2, \cdots, n$.
Thus, the $k n$ hyperplanes
$$
\begin{array}{l}
x_{1}-j_{1}=0\left(1 \leqslant j_{1} \leqslant n\right), \\
x_{2}-j_{2}=0\left(1 \leqslant j_{2} \leqslant n\right), \\
\cdots \cdots . \\
x_{k}-j_{k}=0\left(1 \leqslant j_{k} \leqslant n\right),
\end{array}
$$

clearly meet the requirements.
Hence, the minimum number of hyperplanes that satisfy the conditions is $k n$.