Cardinality of a p-Sylow Let $G$ be a group with cardinal $n=p^{\alpha}m$,  (prime $p$ dividing $n$ and $\gcd(m,p)=1$).
Let $E$ be the set of subsets of $G$ containing $p^{\alpha}$ elements.
I'm trying to understand why $p$ does not divide $\vert E \vert = \binom{p^{\alpha}m}{p^{\alpha}}$
Writing
\begin{equation}
\displaystyle \binom{p^{\alpha}m}{p^{\alpha}}=\frac{p^{\alpha}m}{p^{\alpha}}.\frac{p^{\alpha}m-1}{p^{\alpha}-1}...\frac{p^{\alpha}m-p^{\alpha}+1}{1}
\end{equation}
After simplifications, the remaining quantity is not divisible by $p$.
The first fraction of the binomial coefficient can be simplified by $p^{\alpha}$ but is it true for the other fractions ?
I thank you in advance for any suggestions.
 A: Recall that the $p$-adic valuation of $N$ is $v_p(N)=r$ when $p^r$ is the highest power of $p$ dividing $N$, so that $(p,N)=1$ if and only if $v_p(N)=0$.
By counting powers of $p$ smaller than $n$ we get the formula
$$
v_p(n!)=\lfloor \frac np\rfloor+\lfloor\frac n{p^2}\rfloor+\lfloor\frac n{p^3}\rfloor+\cdots.
$$
Now let $n=p^am$ with $(p,m)=1$. The formula above reads
$$
v_p((p^am)!)=p^{a-1}m+p^{a-2}m+\cdots +m.
$$
Using the same formula again we get
$$
v_p((p^a(m-1))!)=p^{a-1}(m-1)+p^{a-2}(m-1)+\cdots +(m-1)
$$
and
$$
v_p((p^a!)=p^{a-1}+p^{a-2}+\cdots+p.
$$
Now since $v_p(\frac A{BC})=v_p(A)-v_p(B)-v_p(C)$ we can apply the above computation to
$$
v_p(\binom{p^am}{p^a})=v_p((p^am)!)-v^p(p^a!)-v_p((p^a(m-1))!)=0
$$
proving the assertion.
A: Yes, it's true for the other fractions as well. Let $k\in\{1,2,...,p^{\alpha}-1\}$. We can write $k=p^sb$ where $p$ does not divide $b$. Obviously $0\leq s\leq \alpha-1$. Then:
$p^{\alpha}m-k=p^{\alpha}m-p^sb=p^s(p^{\alpha-s}m-b)$
The expression in the brackets is not divisible by $p$, hence $p$ divides $p^{\alpha}m-k$ with multiplicity $s$. Similarly:
$p^{\alpha}-k=p^{\alpha}-p^sb=p^s(p^{\alpha-s}-b)$
So again, $p$ divides $p^{\alpha}-k$ with multiplicty $s$. 
A: Take general term $\frac{p^am-i}{p^a-i}$. Here $i$ is between $\{0,1,\cdots,p^a-1\}$. 
The number $i$ may or may not be divisible by $p$; however, we can always write $i=p^bt$ where $0\le b< a$ and  $p\nmid t$. 
Now the general term becomes $\frac{p^am-p^bt}{p^a-p^bt}$. You can see that $p^b$ cancels and then the numerator and denominator is not divisible by $p$.
Upshot: If, in general term, the numerator is divisible by $p^b$ then so is the denominator and vice-versa. Hence your conclusion!
