Odd! Product!

$\begin{aligned} 1\times 3\times 5\times 7\times 9... \times 9999 & = \frac {1\times {\color{#3D99F6}2} \times 3 \times {\color{#D61F06}4} \times 5 \times{\color{#3D99F6}6} \times ...\times 9999 \times {\color{#D61F06}10000}} {{\color{#3D99F6} 2}\times {\color{#D61F06} 4}\times {\color{#3D99F6} 6} \times {\color{#D61F06} 8} \times {\color{#3D99F6} 10} \times... \times {\color{#3D99F6} 9998} \times \color{#D61F06} {10000}} \\ & = \frac {1\times 2\times 3\times 4\times 5... \times10000} {2\cdot{} 1 \times 2\cdot{} 2 \times 2\cdot{}3 \times 2\cdot{} 4 \times 2\cdot{} 5... \times 2\cdot{} 5000} \\ & = \boxed{\dfrac {10000!}{2^{5000}5000!} } \end{aligned}$

$1 \cdot 3 \cdot 5 \cdots 9999$

$= \dfrac{1 \cdot 2 \cdot 3 \cdot 4 \cdots 10000}{2 \cdot 4 \cdot 6 \cdots 10000}$

$= \dfrac{10000!}{2^{5000} \cdot 1 \cdot 2 \cdot 3 \cdots 5000}$

$= \dfrac{10000!}{2^{5000}\cdot5000!}$

Well I did something else, I tried it for 10. So the positive odd integers less than 10 are 1,3,5,7 and 9 and their product is 945. Then I tried all the possible solutions by replacing 10000! by 10! and 9999! by 9! and so on, so when I calculated 10!/ 2^5 * 5! it gave me 945 so I deduced it will be the same for 10000 ....

I tried formula in the second answer to find the product of all positive odd integers less than 4 and found that it works

This second formula also worked for all positive odd integers less 6, 8,10,12 and 14.

Then I noticed that 1x3x5x7x9x11x13x...x47x49 = (1x 2 x3x 4 x5x 6 x7x...x 50 )/( 2x4x6x8x10x...x50 )

It works for 16,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46,48,52,54,56,58,60,62,64 and 66 so it must work for 10000.

So, 1x3x5x7x...x9999 = (1x2x3x4x5x...x10000)/(2x4x6x8x...10000) =10000!/(2^5000x5000!)