How to prove $\frac{1}{q} \binom{p^k q}{p^k} \equiv 1 \mod p\;\;$? An argument that I just came across asserts that if $p$ is prime, and $k, q\in \mathbb{Z}_{>0}$, then
$$\frac{1}{q} \binom{p^k q}{p^k} \equiv 1 \mod p \;.$$
This assertion was made in a way (i.e. without proof or citation) that suggests the author considers it pretty obvious, but it isn't so to me.  If there's a simple proof of it, I'd like to see it.
(Note: $q$ need not be coprime to $p$.)
(Searching online I found a result called Lucas' Theorem, which looks like it may include the equivalence above as a special case, but I'm not sure yet.  In any case, I did not find the proof of Lucas' theorem particularly easy, but, if the equivalence above is indeed a special case of Lucas' theorem, I hope there may be an easier proof for it.)
 A: Your assertion is as follows.
$${1 \over q} \prod_{i=1}^{p^k}{p^k(q-1)+i \over i} \equiv 1 \mod{p}$$
Let $\mathbb{p}(n)$ be largest $e \in \mathbb{N}_0$ so that $p^e \mid n$ and $\sigma(n) = {n \over p^{\mathbb{p}(n)}}$. It is apparent that for every $i < p^k$ $\mathbb{p}(p^k(q-1)+i) = \mathbb{p}(i)$ and $\sigma(p^k(q-1)+i) \equiv \sigma(i) \mod{p}$. Finally for the last $i = p^k$ we have ${p^k(q-1)+p^k \over p^k} = q$ which is eliminated by the fraction $1 \over q$.
A: In the process of trying to understand P.K.'s I came up with the following proof.  This could very well be exactly P.K.'s proof, although since I can't fully follow P.K.'s argument I can't say for sure.
Clearly
$$
\frac{1}{q} \binom{p^k q}{p^k} = \binom{p^k q - 1}{p^k - 1} = \prod_{i=1}^{p^k - 1}\frac{p^k (q - 1) + i}{i} = \prod_{i=1}^{p^k - 1}\left( \frac{p^k (q - 1)}{i} + 1 \right) \,.$$
(In what follows, $i$ always represents an arbitrary index of the product in the RHS above, or IOW, an arbitrary element of $\{1,\dots,p^k - 1\}$.  Also, every statement about $i$ below should be construed as being implicitly preceded with a "$\forall \, i \in \{1,\dots,p^k - 1\}$".)
The key observation is that $i$ can be factored as $i = p^{k - j_i}c_i$, where $0 < j_i \leq k$ is an integer, and $p \nmid c_i$.  This means that
$$\frac{p^k (q - 1) + i}{i} = p^{j_i}\frac{(q - 1)}{c_i} + 1\,,$$
Now, $p \nmid c_i$ means that there exists an integer $0 < d_i < p$ such that
$$\frac{(q - 1)}{c_i} \equiv d_i\,(q - 1) \mod p\;.$$
Since $j_i > 0$, the congruence above implies that
$$\frac{p^k (q - 1) + i}{i} = p^{j_i}\frac{(q - 1)}{c_i} + 1 \;\;\equiv\;\; p^{j_i} d_i (q - 1) + 1  \;\;\equiv\;\; 1\mod p\,,$$
From this it follows immediately that
$$
\frac{1}{q} \binom{p^k q}{p^k} = 
\prod_{i=1}^{p^k - 1}\left( \frac{p^k (q - 1)}{i} + 1 \right) 
\;\;\equiv\;\; \prod_{i=1}^{p^k - 1} 1 \;\;\equiv 1\mod p\,.$$
