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
$$ \frac{(2n)!}{n! 2^{n}} = \frac{\prod\limits_{k=1}^{2n} k}{\prod\limits_{k=1}^{n} (2k)} = \prod_{k=1}^{n} (2k-1) \in \Bbb{Z}. $$
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.