How do you simplify $\dfrac{(2n)!}{n!}$?

Hint: A factorial of a number multiplied by all of the integers preceding it. Example: “Five Factorial” = $5!=5\times 4\times 3\times 2\times 1=120$. Hence factorial of a number $n$ is the product of all the numbers from $1$ to $n$. So we have
$n!=n\cdot (n-1)\cdot (n-2)\cdot ...\cdot 2\cdot 1$
Although there is no simplification of $\dfrac{(2n)!}{n!}$, but there are other ways of expressing it, such as
$\dfrac{(2n)!}{n!}=\prod\limits_{k=0}^{n-1}{(2n-k)=(2n)(2n-1)...(n+1)}$ .

Complete step by step solution:
To simplify: $\dfrac{(2n)!}{n!}$
By the definition of factorial function,
$(2n)!=(2n)(2n-1)...2\cdot 1$
And $n!=n(n-1)(n-2)...\cdot 2\cdot 1$
Now putting the value of $(2n)!$ in $\dfrac{(2n)!}{n!}$ , we get
$\dfrac{(2n)!}{n!}=\dfrac{1}{n!}(1\cdot 2\cdot 3\cdot ...\cdot 2n)$
This can also be written as:
$\begin{align}
  & \dfrac{(2n)!}{n!}=\dfrac{1}{n!}(2\cdot 4\cdot 6\cdot ...\cdot 2n)(1\cdot 3\cdot 5\cdot ...\cdot (2n-1)) \\
 & =\dfrac{1}{n!}(1\cdot 2)(2\cdot 2)(3\cdot 2)...(n\cdot 2)(1\cdot 3\cdot 5\cdot ...(2n-1)) \\
 & =\dfrac{1}{n!}(1\cdot 2\cdot 3\cdot ...n){{2}^{n}}(1\cdot 3\cdot 5\cdot ...\cdot (2n-1)) \\
 & =\dfrac{1}{n!}n!\cdot {{2}^{n}}(1\cdot 3\cdot 5\cdot ...\cdot (2n-1)) \\
 & ={{2}^{n}}(1\cdot 3\cdot 5\cdot ...\cdot (2n-1)) \\
\end{align}$

Therefore, $\dfrac{(2n)!}{n!}={{2}^{n}}(1\cdot 3\cdot 5\cdot ...\cdot (2n-1))$.

Note: Once we know the value of n which can be either positive, negative, or zero, we can easily find the value of $\dfrac{(2n)!}{n!}$ by putting n in the solution. Factorials have a usage in various areas of mathematics such as probability, combinations and permutations, Taylor’s series, exponential functions, Binomial expansion, and many more.