Playing with large numbers

From Wilson theorem we can have following $1000002 !\equiv -1\mod(1000003)$ and $(-6)\times 999999! \equiv -1\mod(1000003)$ Now we wish to find the modulo inverse of $6\mod(1000003)$ . Since $1000003 =6(166667)+1$ Therefore, $999999!\equiv -166667\mod(1000003)\implies 999999!\equiv 833336\mod(1000003)$ So, $\boxed{833336}$ is the required answer.