Let $F$ be a field of characteristic $p$. For each natural number $n$, denote
$$
n=1+1+\cdots+1 \quad \text { ( } n \text { summands). }
$$
In particular, in $F$, we have $p=0, p+1=1$, etc.
Verify that for all $n=0,1, \ldots, p(=0)$, we can define binomial coeficients in $F$ in the usual manner:
$$
\left(\begin{array}{l}
n \\
k
\end{array}\right)=\frac{n(n-1) \cdots(n-k+1)}{k(k-1) \cdots 1}=n(n-1) \cdots(n-k+1)[k(k-1) \cdots 1]^{-1}
$$
if $0<k<p$, and
$$
\left(\begin{array}{l}
n \\
0
\end{array}\right)=1, \quad\left(\begin{array}{l}
p \\
p
\end{array}\right)=1 .
$$
Verify that the following well-known formula is valid:
$$
\left(\begin{array}{l}
n \\
k
\end{array}\right)+\left(\begin{array}{c}
n \\
k-1
\end{array}\right)=\left(\begin{array}{c}
n+1 \\
k
\end{array}\right) .
$$