Q.Show that $\dfrac{(2n)!}{n!} = 2^n\,\{1\cdot3\cdot5\cdots(2n-1)\}$

For a natural number $n$, the factorial of $n$, written $n!$, is the product of all natural numbers from 1 up to $n$: $n! = 1\times2\times3\times\cdots\times(n-1)\times n$. It can equally be read in the reverse (and often more useful) order, $n!=n\times(n-1)\times(n-2)\times\cdots\times2\times1$, which is what makes properties like $n!=n\times(n-1)!$ so natural: the factorial of a number is that number times the factorial of the number just below it. By convention, $0!$ is defined to equal $1$ (even though 0 is not itself a natural number), which keeps formulas like ${}^nC_0=1$ and ${}^nP_0=1$ consistent without needing a special-case exception. Factorial notation grows extremely fast ($10!$ is already over 3.6 million), so most factorial computations in practice are done by CANCELLING a shared tail between a numerator and a denominator (writing the larger factorial as extra factors times the smaller one) rather than by expanding both fully — this is the technique behind simplifying expressions like $\dfrac{n!}{(n-r)!}$ …