Lucas' Theorem

Theorem
Let $$p$$ be a prime number.

Let $$n, k \in \Z$$.

Then:
 * $$\binom n k \equiv \binom {\left \lfloor {n / p} \right \rfloor} {\left \lfloor {k / p} \right \rfloor} \binom {n \, \bmod \, p} {k \, \bmod \, p} \pmod p$$

where:
 * $$\binom n k$$ is a binomial coefficient;
 * $$\left \lfloor \cdot \right \rfloor$$ is the floor function.

Hence, if the representation of $n$ and $k$ to the base $p$ are given by:
 * $$n = a_r p^r + \cdots + a_1 p + a_0$$
 * $$k = b_r p^r + \cdots + b_1 p + b_0$$

then:
 * $$\binom n k \equiv \prod_{j=0}^r \binom {a_j} {b_j} \pmod p$$

Proof of first part
First we show that:
 * $$\binom n k \equiv \binom {\left \lfloor {n / p} \right \rfloor} {\left \lfloor {k / p} \right \rfloor} \binom {n \, \bmod \, p} {k \, \bmod \, p} \pmod p$$.

Consider $$\binom n k$$ as the fraction $$\frac {n \left({n-1}\right) \left({n-2}\right) \cdots \left({n-k+1}\right)} {k \left({k-1}\right) \left({k-2}\right) \cdots 1}$$.

This can be expressed as:
 * $$(1) \qquad \binom n k = \left({\frac {n}{k}}\right) \left({\frac {n-1}{k-1}}\right) \left({\frac {n-2}{k-2}}\right) \cdots \left({\frac {n-k+1}{1}}\right)$$.

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 we deal 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:
 * $$\left({\frac {u p + v}{s p + t}}\right) \left({\frac {u p + v-1}{s p + t-1}}\right) \cdots \left({\frac {u p + v-t+1}{s p + 1}}\right) \equiv \left({\frac {v}{t}}\right) \left({\frac {v-1}{t-1}}\right) \cdots \left({\frac {v-t+1}{1}}\right) \pmod p$$

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

These differ by multiples of $$p$$.

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

These fall into $$\left \lfloor {k / p} \right \rfloor$$ 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 $$\left({p-1}\right)!$$ 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:
 * $$\binom{\left \lfloor {\left({n - k \, \bmod \, p}\right) / p} \right \rfloor} {\left \lfloor {k / p} \right \rfloor}$$

Now, if $$k \, \bmod \, p \le n \, \bmod \, p$$, this equals:
 * $$\binom {\left \lfloor {n / p} \right \rfloor} {\left \lfloor {k / p} \right \rfloor}$$

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

is zero.

So the formula holds in general.

Proof of second part
Now we consider the representation of $n$ and $k$ to the base $p$:
 * $$n = a_r p^r + \cdots + a_1 p + a_0$$
 * $$k = b_r p^r + \cdots + b_1 p + b_0$$

Let $$n_1 = \left \lfloor {n / p} \right \rfloor, k_1 = \left \lfloor {k / p} \right \rfloor$$.

We have that:
 * $$n \, \bmod \, p = a_0, k \, \bmod \, p = b_0$$;
 * $$n_1 = a_r p^{r-1} + a_{r-1} p^{r-2} \cdots + a_1, k_1 = b_{r-1} p^{r-2} \cdots + b_1$$.

From the proof of the first part, this gives us:
 * $$\binom n k \equiv \binom {n_1} {k_1} \binom {a_0} {b_0} \pmod p$$

Now we do the same again to the representation to the base $p$ of $$n_1$$ and $$n_2$$.

Thus:
 * $$\binom n k \equiv \binom {\left \lfloor {n_1 / p} \right \rfloor} {\left \lfloor {k_1 / p} \right \rfloor} \binom {a_1} {b_1} \binom {a_0} {b_0} \pmod p$$

And so on until $$\left \lfloor {n_r / p} \right \rfloor$$ and $$\left \lfloor {k_r / p} \right \rfloor$$.

Hence the result.

It appears in his 1878 paper Théorie des Fonctions Numériques Simplement Périodiques in the American Journal of Mathematics Volume 1, issues 2 - 4.