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.

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 $s$. 
