$p^t$ divides $C(n,p^r)$, $p^{t+1}$ doesn't I'm having trouble seeing why this (it's stated as obvious in a proof of Sylow's first theorem): if $n=mp^r$ ($p$ is prime), and $p^t\mid m$ but $p^{t+1}\nmid m$, then $p^t\mid\binom{n}{p^r}$ but $p^{t+1}\nmid\binom{n}{p^r}$.
I also wonder if a combinatorial proof of this fact can be given.
Thank you in advance for any help given.
 A: If we expand ${ mp^r \choose p^r }$ in the usual way we obtain
$${ mp^r \choose p^r } = m \prod_{k=1}^{p^r-1} \frac{mp^r-k}{p^r-k}$$
and since $mp^r -k= (m-1)p^r + (p^r-k)$ the power of $p$ dividing $mp^r-k$ is the same as the power of $p$ dividing $p^r-k.$ Hence the powers of $p$ in the numerator and denominator of the above product cancel, which gives us the result due to the $m$ before the product sign.
A: This is the same idea which I.N.Herstein uses to prove the First Sylow theorem. So i would recommend you to have a peep into his, Topics in Algebra book, Sec 2.12, Group Theory, Page 92

A: As others have mentioned, this follows easily by aligning the numerator and denominator factors modulo $\rm\ p^r \ $ in order to obtain maximal cancellation of powers of $\rm\:p\:.\:$ This is a special case of the simple arithmetical method I described in my post here to prove that binomial coefficients are integers. See also this very simple example for motivation. I think that you'll find that the key conceptual idea is clearer after studying these further examples.
