Proof that $\frac{(2n)!}{2^n}$ is integer I am trying to prove that $\dfrac{(2n)!}{2^n}$  is integer. So I have tried it by induction, I have took  $n=1$, for which we would have $2/2=1$ is integer. So for $n=k$ it is true, so now comes time to proof it for $k+1$, $(2(n+1))!=(2n+2)!$, which is equal to $$1 \times 2 \times 3 \times \cdots \times (2n) \times (2n+1) \times (2n+2),$$ second term would be $$2^{n+1}=2 \times 2^n$$
Finally if we divide $(1 \times 2 \times 3 \times \cdots \times (2n) \times (2n+1) \times (2n+2)$ by $2^{n+1}=2 \times 2^n$,and consider that,$(2n)!/(2^n)$  is integer, we get $(2n+1) \times (2n+2)/2=(2n+1) \times 2 \times (n+1)/2$, we can cancel out $2$, we get $(2n+1)(n+1)$ which is definitely integer.
I am curious this so simple? Does it means that I have proved correctly?

From 1 to $2n$ there are exactly $n$ even numbers. Hence the product $1\cdots 2n=(2n)!$ is divisible by $2\cdot2\cdots 2 \hbox{ ($n$ times)}= 2^n$.
