We are working in the base $2$ world now. First, we need to find the smallest $n$ such that $C_n^8 \ge 1000$. This count is the total numbers containing exactly $8$ ones in a $n$-digit number written in base $2$, including those starting with $0$s. Trying a few candidates find $C_{12}^8=495 < 1000$ and $C_{13}^8=1287 > 1000$. Therefore, the $1000^{th}$ number must have $13$ digits. The highest bit (i.e. the $13^{th}$ digit) is $1$. Fixing this, we need to fill the remaining $12$ digits with exactly $7$ ones. We are looking for the $1000-495=505^{th}$ such number.

Because $C_{12}^7=792 > 505$ but $C_{11}^7=330 < 505$, we know the $12^{th}$ bit is also $1$. Fixing this, we need to fill the remaining $11$ digits with exactly $6$ ones. We are looking for the $505-330=175^{th}$ such number.

Because $C_{11}^6=446 > 175$, $C_{10}^6=210$, and $C_9^6=84$, we conclude the $11^{th}$ digit is $0$ and the $10^{th}$ digit is $1$. This leaves us with $9$ digits to be filled with $5$ ones. We need to find the $175-84=91^{th}$ such number.

Because $C_9^5=126$, and $C_8^5=56$, therefore we find the $9^{th}$ digit is $1$. Then, we need to fill the remaining $8$ digits with $4$ ones and looking for the $91-56=35^{th}$ such number.

Because $C_8^4=70$ and $C_7^4=35$, so we know the number we are looking for is the largest $7$-digit number with $4$ ones, i.e. $1111000_{(2)}$. Therefore the answer is $$1101101111000_{(2)}=\boxed{7032}_{(10)}$$