How to prove ${{pm} \choose {pn}}\equiv{m \choose n} \pmod{p}$. Question:(1) if p is a prime and m,n $\in$ N,prove that ${{pm} \choose {pn}}\equiv{m \choose n} \pmod p$ (the book gives me a hint: think about $(1+x)^{pm}$ and $(1+x^m)^p$ in $F_{p}(x)$.
(2) Prove that ${{p^r}m \choose {p^r}n}\equiv{m \choose n}$(mod p) for all interger r>0.
I am stuck in this question, since this question is in the abstract algebra book, so I think we should need to use abstract algebra knowledge to solve it. But I still don't know how to prove this question. Can someone help me solve this problem, or give me some hints? As for the part(2), I am trying to use mathematical induction to prove it, but it looks hard.
 A: Algebraic/Combinatorial Proof:
There are $p^rm$ objects, each having one of the available $m$ colors, and for each of the $m$ colors, there are $p^r$ objects with that color and they are indistinguishable.  We want to arrange these objects on $n$ arrays, each having size $p^r$, and let $S$ be the set of all such arrangements.  Obviously, $|S|=\binom{p^rm}{p^rn}$.  The group $G:=C_{p^r}^n$, where $C_k$ is the cyclic group of order $k$, acts on $S$ by circularly shifting the entries of the $n$ arrays.
Let $\mathcal{O}$ be the set of all orbits $O$ of the action of $G$ on $S$ with $|O|>1$.  The class equation of the action of $G$ on $S$ is given by
$$|S|=\left|S_0\right|+\sum_{O\in\mathcal{O}}\,\left|O\right|\,.$$
Note that elements of $S_0$ are simply $n$ arrays each of whose elements have the same color.  That is, $\left|S_0\right|=\binom{m}{n}$.  Note also that each orbit $O$ in $\mathcal{O}$ is of size $p^{k(O)}$ for some integer $k(O)\geq 2$ (because, if there is an array with more than one colors, then there are at least two arrays with more than one colors).  Hence,
$$\binom{p^rm}{p^rn}=\binom{m}{n}+p^2\,\sum_{O\in\mathcal{O}}\,p^{k(O)-2}\equiv \binom{m}{n}\pmod{p^2}\,.$$
A: HINT: For the first problem first show that in $\Bbb F_p(x)$ we must have
$$(1+x^m)^p=\sum_k\binom{p}kx^{km}=1+x^{pm}\;;$$
this uses the fact that $p$ is prime. Then use this to show that $(1+x)^p=1+x^p$ and hence that
$$\sum_k\binom{pm}kx^k=(1+x)^{pm}=(1+x^p)^m=\sum_k\binom{m}kx^{pk}\;.$$
Finally, equate coefficients.
For the second, note that $p^rm=p(p^{r-1}m)$, and use induction and the first problem.
