$4.1992^{1002}$ when divided by 1993 leaves a remainder of
$$
\begin{aligned}
4 \cdot 1992^{1002} & =(1993-1)^{2 \cdot 2 \cdot 2 \cdot 3 \cdot 83} \\
& =\left(1993^{2}-2 \cdot 1993+1\right)^{2 \cdot 2 \cdot 3 \cdot 83}
\end{aligned}
$$

Continuing the calculation in a similar manner, the first term that appears is 1993, and the last term is 1. Therefore, the remainder when $1992^{1992}$ is divided by 1993 is 1.