Wolfram Alpha says that this result is true:
$$\frac{\Gamma(n+1)}{\Gamma(\frac{n}{2}+1)} = \frac{\Gamma(\frac{n}{2} +\frac{1}{2})}{\Gamma(\frac{1}{2})} \times 2^n$$
This implies a curious result for even $n$; the product of the numbers from $\frac{n}{2}+1$ to $n$ is equal to $2^{n/2}$ times the product of the odd numbers from $1$ to $n-1$.
For example, if we split the naturals from $1$ to $10$ into two groups:
$$10 \:\:\:9\:\:\:8\:\:\:7\:\:\:6\:\:\: | \:\:\:5 \:\:\:4 \:\:\:3\:\:\: 2\:\:\: 1$$
And then subtract 0.5 from every number in the second group:
$$10 \:\:\:9\:\:\:8\:\:\:7\:\:\:6\:\:\: | \:\:\:4.5 \:\:\:3.5 \:\:\:2.5\:\:\: 1.5\:\:\: 0.5$$
Then the identity tells us that the product of the first group is exactly equal to $2^n$ times the product of the second group, where $n=10$ in this example. My question is this:
Is there an intuitive reason why this number is $2^n?$
For example, if we multiplied the second group by $2^{n/2}$ then we would have
$$\:\:\:9 \:\:\:7 \:\:\:5\:\:\: 3\:\:\: 1$$
I would like to have an intuitive reason as to why multiplying this by another $2^{n/2}$ makes it equivalent to the group on the left.
This is because, using factorial notation,
$$(2n-1)\times(2n-3)\times\cdots\times1=\frac{(2n)!}{n!2^n}$$ and thus $$2^n(2n-1)\times(2n-3)\times\cdots\times1=\frac{(2n)!}{n!}=2n\times(2n-1)\times\cdots\times(n+1)$$