A direct proof of $\binom{m\,p^k-1}{p^k-1}\equiv1~(\text{mod $p$})$? In Nathan Jacobson's "Basic algebra I" the exercises 1.13.11-14
prove the following extension of
(a part of) the Sylow's second theorem:

If $p$ is a prime and $p^k\bigm||G|$,
  then  the number of subgroups of $G$ of order $p^k$ is congruent $1~(\text{mod $p$})$.

A side product of these exercises is the following result in number theory:
if $p$ is a prime, $m>0$ is an integer, and $k\geq 0$ is an integer,
then
$$
\frac{1}{m}\!\binom{m\cdot p^k}{p^k} \:\equiv\: 1~\:(\text{mod $p$})~.
$$
The left hand side of the congruence can be rewritten as $\binom{m\,p^k-1}{p^k-1}$,
so we have the congruence
$$
\binom{m\cdot p^k-1}{p^k-1} \:\equiv\:1~\:(\text{mod $p$})~, \tag{1}
$$
which happens to be true for every integer $m$
(and a prime $p$ and an integer $k\geq 0$).
Can you prove this congruence directly, using only number-theoretic reasoning,
without slyly engaging properties of groups in some disguise or other?
I have tried (and tried$\ldots$), and failed.
 A: Following up the hints.
Lucas' theorem:

If $m$ and $n$ are non-negative integers and $p$ is a prime, then
  $$
\binom{m}{n} \:\equiv\: \prod_{i=0}^k \binom{m_i}{n_i}~~(\text{mod $p$})~,
$$
  where
  $$
\begin{aligned}
m ~&\:=\: m_kp^k + m_{k-1}p^{k-1} + \cdots + m_1p + m_0~, \\
n ~&\:=\: n_kp^k + n_{k-1}p^{k-1} + \cdots + n_1p + n_0
\end{aligned}
$$
  are the base $p$ expansions of $m$ and $n$.

Note that in the base $p$ expansions we can have $m_k=0$ or $n_k=0$ or both;
increasing the $k$ just adds factors $\binom{0}{0}=1$
to the product on the right hand side of the Lucas' congruence.
How we apply Lucas' theorem in our case?
We can assume that $m>0$; if congruence $(1)$ holds for all positive integers $m$,
then it holds for all integers $m$
(actually, if it holds for any $p\cdot(p^k-1)!$ consecutive integers $m$,
then it holds for all integers $m$).
Also, we can assume that $k>0$, because $(1)$ is trivially true when $k=0$.
First,
$$
p^k-1 \:=\: (p-1)p^{k-1} + \cdots + (p-1)p + (p-1)
$$
(all digits of the expansion are $p-1$).
Next, since $m\cdot p^k-1\geq p^k-1$
and clearly $m\cdot p^k - 1\equiv p^k - 1~(\text{mod $p^k$})$,
we have the base $p$ expansion, for some $l\geq k$,
$$
m\cdot p^k-1 \:=\: m_l p^l + \cdots + m_k p^k + (p-1)p^{k-1} + \cdots + (p-1)p + (p-1)~.
$$
Now Lucas' theorem tells us that
$$
\binom{m\cdot p^k-1}{p^k-1}
    \:\equiv\: \binom{m_l}{0}\cdots\binom{m_k}{0}\binom{p-1}{p-1}\cdots\binom{p-1}{p-1}
    \:\equiv\: 1~~(\text{mod $p$})~.
$$
(Did I say thank you for the hints?)
A: In the field $\mathbb F_p(x)$,
$$(1-x)^{mp^k-1} = \frac{(1-x)^{mp^k}}{1-x} = \frac{(1-x^{p^k})^{m}}{1-x} = \frac{1-x^{p^k}}{1-x} (1-x^{p^k})^{m-1} = (1+x+\dots+x^{p^k-1}) (1-x^{p^k})^{m-1}$$
Now the coefficient of $x^{p^k-1}$ in $(1+x+\dots+x^{p^k-1}) (1-x^{p^k})^{m-1}$ is equal to the coefficient of $x^{p^k-1}$ in $1+x+\dots+x^{p^k-1}$ because the other factor $(1-x^{p^k})^{m-1}$ is a polynomial in $x^{p^k}$ with constant coefficient $1$.
Therefore ${mp^k-1 \choose p^k-1} = 1 \mod p$.
