Inspired by Pi Han

For this product to be odd, we will require that each and every coefficient $\binom{n}{k}$ be odd. Now by a Lucas' Theorem , we are thus looking for the largest $n \lt 10000$ which has a base $2$ expansion that consists only of $1$ 's. Since $2^{13} = 8192$ is the highest power of $2$ less than $10000$ we know then that $8192 - 1 = \boxed{8191}$ is the value of $n$ we are looking for, (as its base $2$ expansion consists of only $1$ 's, and any value $8192 \le n \le (2^{14} - 2) = 16382$ will have at least one $0$ in its base $2$ expansion).

(Odd numbers are coloured in)

$\displaystyle \prod_{k=0}^{n} \binom{n}{k}$ is the product of all the numbers in a row of Pascal's triangle. In order for the product to be odd, all numbers in that row must be odd. Looking at the picture, this occurs in the $2^{a} - 1$ th rows.

The highest power of $2$ smaller than $10000$ is $2^{13} = 8192$ . The row before that is $8192 - 1 = 8191$ , so $n = 8191$ .