Yes. There is also a positive integer $ k$ such that none of the digits $ 2,3,4,5,6,7,8,9$ appear in the decimal representation of the number $ 2002!\cdot k$ because it has the form $ 11...100...0$.

Please show the proof..

Well, in the sequence $ 1, 11, 111, 1111, ...$ there are two numbers $ k, l$ such that $ k\equiv l(mod2002!)$. Their difference is divisible by $ 2002!$ and has the form $ 11...100...0$.