Let $n=p^rm$, where $p$ is a prime not dividing an integer $m\ge 1$. Prove that $p$ is a prime not dividing $C_n^{p^r}$ I    have some problems solving the question, it appears in my textbook. if you can solve the question,please tell me. I am glad to know how to solve the question. Thank you for your help.

Let $n=p^rm$, where $p$ is a prime not dividing an integer $m\ge 1$. Prove that $p$ is a prime not dividing $C_n^{p^r}$.

 A: This is a special case of Lucas' theorem.
Compute in the polynomial ring $\mathbb{Z}_{p}[x]$:
$$
(1+x)^{p^{r} m} = (1 + x^{p^{r}})^{m} = 1 + m x^{p^{r}} + \dots
$$
so that by the binomial theorem
$$
\binom{p^{r} m}{p^r} \equiv m \not\equiv 0\pmod{p}.
$$
Here we have used the fact that in $\mathbb{Z}_{p}[y]$ one has
$$
(1 + y)^{p} = 1 + y^{p},
$$
that is,
$$
\binom{p}{i} \equiv 0 \pmod{p}, \quad\text{for $0 < i < p$.}
$$
A: Check, that
$$\binom{n}{p^r}=\prod_{i=0}^{p^r-1}\frac{n-i}{p^r-i}$$
We want to show, that none of the factors is divisible by $p$, so take $i\leq p^r-1$ arbitrarily.
Let $k$ be the highest power of $p$ dividing $i$ (i.e. $p^k\mid i\wedge p^{k+1}\nmid i$). Cleary $k<r$, hence we have the  divisibility chain $p^k\mid p^{k+1}\mid p^r\mid n$, which yields: 


*

*$p^k\mid n-i$ and $p^k\mid p^r-i$

*Assume $p^{k+1}\mid n-i$ or $p^{k+1}\mid p^r-i$. As $p^{k+1}\mid n$ and $p^{k+1}\mid p^r$, in both cases this would imply $p^{k+1}\mid i$, contrary to the assumption, that $k$ is maximal.


So, in the $i$-th factor of the above product, $p$ occurs exactly $k$ times both in the denominator and in the nominator, they cancel out, and the factor is not divisible by $p$.
A: the power of $p$ dividing $n!$ is $n-\sigma_p(n)$ where $\sigma(k)$ is the sum of the binary digits in the $p$-ary notation for $k$. if
$$
B= \binom{n}{p^r} = \frac{(mp^r)!}{(p^r)!((m-1)p^r)!}
$$
then the power of $p$ dividing $B$ is (subtracting two denominator terms from the numerator term)
$$
mp^r - \sigma(mp^r) - ( p^r -\sigma(p^r)) - ( (m-1)p^r -\sigma((m-1)p^r)) \\
= \sigma(p^r) + \sigma((m-1)p^r) - \sigma(mp^r) \\
= 1 + \sigma(m-1) - \sigma(m) \\
= 0
$$
because the least significant digit of $m$ is non-zero.
