Lucas' Theorem

Theorem
Let $p$ be a prime number.

Let $n, k \in \Z_{\ge 0}$.

Then:
 * $\dbinom n k \equiv \dbinom {\floor {n / p} } {\floor {k / p} } \dbinom {n \bmod p} {k \bmod p} \pmod p$

where:
 * $\dbinom n k$ denotes a binomial coefficient
 * $n \bmod p$ denotes the modulo operation
 * $\floor \cdot$ denotes the floor function.

Proof
First we show that:
 * $\dbinom n k \equiv \dbinom {\floor {n / p} } {\floor {k / p} } \dbinom {n \bmod p} {k \bmod p} \pmod p$

Consider $\dbinom n k$ as the fraction:
 * $\dfrac {n \paren {n - 1} \paren {n - 2} \cdots \paren {n - k + 1} } {k \paren {k - 1} \paren {k - 2} \cdots 1}$

This can be expressed as:
 * $(1): \quad \dbinom n k = \paren {\dfrac n k} \paren {\dfrac {n - 1} {k - 1} } \paren {\dfrac {n - 2} {k - 2} } \cdots \paren {\dfrac {n - k + 1} 1}$

Let $k = s p + t$ from the Division Theorem.

Thus:
 * $t = k \bmod p$

The denominators of the first $t$ factors in $(1)$ do not have $p$ as a divisor.

Now let $n = u p + v$, again from the Division Theorem.

Thus:
 * $v = n \bmod p$

Now, when dealing with non-multiples of $p$, we can work modulo $p$ in both the numerator and denominator, from Common Factor Cancelling in Congruence.

So we consider the first $t$ factors of $(1)$ modulo $p$.

These are:
 * $\paren {\dfrac {u p + v} {s p + t} } \paren {\dfrac {u p + v - 1} {s p + t - 1} } \cdots \paren {\dfrac {u p + v - t + 1} {s p + 1} } \equiv \paren {\dfrac v t} \paren {\dfrac {v - 1} {t - 1} } \cdots \paren {\dfrac {v - t + 1} 1} \pmod p$

So, these first $t$ terms of $(1)$ taken together are congruent modulo $p$ to the corresponding terms of:
 * $\dbinom {n \bmod p} {k \bmod p}$

These differ by multiples of $p$.

So we are left with $k - k \bmod p$ factors.

These fall into $\floor {k / p}$ groups, each of which has $p$ consecutive values.

Each of these groups contains exactly one multiple of $p$.

The other $p - 1$ factors in a given group are congruent (modulo $p$) to $\paren {p - 1}!$ so they cancel out in numerator and denominator.

We now need to investigate the $\left \lfloor {k / p} \right \rfloor$ multiples of $p$ in the numerator and denominator.

We divide each of them by $p$ and we are left with the binomial coefficient:
 * $\dbinom {\floor {\paren {n - k \bmod p} / p} } {\floor {k / p} }$

Now, if $k \bmod p \le n \bmod p$, this equals:
 * $\dbinom {\floor {n / p} } {\floor {k / p} }$

Otherwise, if $k \bmod p > n \bmod p$, the other factor:
 * $\dbinom {n \bmod p} {k \bmod p}$

is zero.

So the formula holds in general.