Prove that $\frac{100!}{50!\cdot2^{50}} \in \Bbb{Z}$ I'm trying to prove that : 
$$\frac{100!}{50!\cdot2^{50}}$$
is an integer .
For the moment I did the following : 
$$\frac{100!}{50!\cdot2^{50}} = \frac{51 \cdot 52 \cdots 99 \cdot 100}{2^{50}}$$
But it still doesn't quite work out . 
Hints anyone ? 
Thanks 

for the number of multiples of $2$:
$$100 = 52 + 2(n_1 - 1)$$
$$n_1 = 25$$
for the number of multiples of $4$:
$$100 = 52 + 4(n_2 - 1)$$
$$n_2 = 13$$
for the number of multiples of $8$:
$$96 = 56 + 8(n_3 - 1)$$
$$n_3 = 6$$
for the number of multiples of $16$:
$$96 = 64 + 16(n_4 - 1)$$
$$n_4 = 3$$
for the number of multiples of $32$:
$$96 = 64 + 32(n_5 - 1)$$
$$n_5 = 2$$
for the number of multiples of $64$:
$$64 = 64 + 64(n_6 - 1)$$
$$n_6 = 1$$
To get the number of $2$'s as factors, add all the $n$'s up, which yields $50$.
This should cancel the $2^{50}$ at the denominator, and thus prove that the expression is an integer.
