Let M be a set with m elements, and divide it into m_i cycles of length pⁱ for the various values of i. Then each of these cycles can be rotated separately, so that a group G which is the Cartesian product of cyclic groups C_pⁱ acts on M. It thus also acts on subsets N of size n. Since the number of elements in G is a power of p, the same is true of any of its orbits. Thus in order to compute ${\displaystyle {\tbinom {m}{n}}}$ modulo p, we only need to consider fixed points of this group action. The fixed points are those subsets N that are a union of some of the cycles. More precisely one can show by induction on k-i, that N must have exactly n_i cycles of size pⁱ. Thus the number of choices for N is exactly ${\displaystyle \prod _{i=0}^{k}{\binom {m_{i}}{n_{i}}}{\pmod {p}}}$.

This proof is due to Nathan Fine.^[2]

If p is a prime and n is an integer with 1 ≤ n ≤ p − 1, then the numerator of the binomial coefficient

${\displaystyle {\binom {p}{n}}={\frac {p\cdot (p-1)\cdots (p-n+1)}{n\cdot (n-1)\cdots 1}}}$

is divisible by p but the denominator is not. Hence p divides ${\displaystyle {\tbinom {p}{n}}}$. In terms of ordinary generating functions, this means that

${\displaystyle (1+X)^{p}\equiv 1+X^{p}{\pmod {p}}.}$

Continuing by induction, we have for every nonnegative integer i that

${\displaystyle (1+X)^{p^{i}}\equiv 1+X^{p^{i}}{\pmod {p}}.}$

Now let m be a nonnegative integer, and let p be a prime. Write m in base p, so that ${\displaystyle m=\sum _{i=0}^{k}m_{i}p^{i}}$ for some nonnegative integer k and integers m_i with 0 ≤ m_i ≤ p-1. Then

${\displaystyle {\begin{aligned}\sum _{n=0}^{m}{\binom {m}{n}}X^{n}&=(1+X)^{m}=\prod _{i=0}^{k}\left((1+X)^{p^{i}}\right)^{m_{i}}\\&\equiv \prod _{i=0}^{k}\left(1+X^{p^{i}}\right)^{m_{i}}=\prod _{i=0}^{k}\left(\sum _{n_{i}=0}^{m_{i}}{\binom {m_{i}}{n_{i}}}X^{n_{i}p^{i}}\right)\\&=\prod _{i=0}^{k}\left(\sum _{n_{i}=0}^{p-1}{\binom {m_{i}}{n_{i}}}X^{n_{i}p^{i}}\right)=\sum _{n=0}^{m}\left(\prod _{i=0}^{k}{\binom {m_{i}}{n_{i}}}\right)X^{n}{\pmod {p}},\end{aligned}}}$

where in the final product, n_i is the ith digit in the base p representation of n. This proves Lucas's theorem.