Notation/simplification: $(n-1)(n-3)(n-5)...(3)(1)$ How would one write this in simplified form? I am aware that the answer is $\frac{n!}{2^{n/2}(n/2)!}$. How to arrive at this answer?
 A: The number $n$ is requiered to be even.
$$1 \cdot 3 \cdot 5 \cdot \ldots \cdot (n-5)(n-3)(n-1) \\[1em]
=\frac{1 \cdot \color{blue}{2} \cdot 3 \cdot \color{blue}{4} \cdot 5 \cdot \color{blue}{6} \cdot \ldots \cdot (n-5) \color{blue}{(n-4)} (n-3) \color{blue}{(n-2)} (n-1) \color{blue}{n} }{\color{blue}{2 \cdot 4 \cdot 6 \cdot \ldots \cdot (n-4) (n-2) n} }$$
Then factor a $2$ out of each one of the $\tfrac{n}2$ even numbers in the denominator:
$$=\frac{n!}{ 2^{n/2} \big(1 \cdot 2 \cdot 3 \cdot \ldots \cdot (\tfrac{n}2-2) (\tfrac{n}2-1) \tfrac{n}2\big)} \\[1em]
=\frac{n!}{2^{n/2} (\tfrac{n}2)!}$$
A: Assuming $n$ is even, then
$$
\require{cancel}
\begin{align}
\frac{n!}{2^{n/2}(n/2)!}
&=\frac{n(n-1)(n-2)(n-3)\cdots(2)(1)}{2^{n/2}(n/2)(n/2-1)(n/2-2)\cdots1}
\\&=\frac{n(n-1){(n-2)}(n-3)\cdots{(2)}(1)}{{(n)}{(n-2)}{(n-4)}\cdots{2}}
\\&=\frac{\cancel n(n-1)\cancel{(n-2)}(n-3)\cdots\cancel{(2)}(1)}{\cancel{(n)}\cancel{(n-2)}\cancel{(n-4)}\cdots\cancel{2}}
\end{align}
$$
In the second step, the twos in $2^{n/2}$ were each pulled into one of the factors in the denominator. 
