<s>[INST] Prove by induction: for $n \ge 0$, $\frac{(2n)!}{n!2^n}$ is an integer Another prove by induction question:
for $n \ge 0$, $$\frac{(2n)!}{n!2^n}$$ is an integer
Base step: 
$$n = 0$$
$$\frac{(2 \times 0)!}{0! \times 2^0} = \frac{0!}{1 \times 1} = 1$$
Induction step: please help
 [/INST] Proof:: (Without Using induction::) Let $$Z=\frac{(2n)!}{n!\cdot 2^n} = \frac{(2n)\cdot (2n-1)\cdot (2n-2)\cdots4\cdot 3 \cdot 2 \cdot 1}{n!\cdot 2^n}$$
so $$Z = \frac{\underbrace{(2n)\cdot (2n-2)\cdot (2n-4)\cdots4\cdot 2}\times \underbrace{(2n-1)\cdot (2n-3)\cdot (2n-5)\cdots3\cdot 1}}{n!\cdot 2^n}$$ 
So $$Z = \frac{2^n\cdot \left[n\cdot (n-1)\cdot (n-2)\cdots 4\cdot 2\right]\times \underbrace{(2n-1)\cdot (2n-3)\cdot (2n-5)\cdots3\cdot 1}}{n!\cdot 2^n}$$
So $$Z = \frac{2^n\cdot n!\times \underbrace{(2n-1)\cdot (2n-3)\cdot (2n-5)\cdots3\cdot 1}}{2^n\cdot n!}$$
So $$Z = \underbrace{(2n-1)\cdot (2n-3)\cdot (2n-5)\cdots3\cdot 1}$$ is an Integer.
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