4. The last digit of $1+2^{2}+3^{3}+4^{4}+\cdots+1992^{1992}$ is ( ).
A. 8
B. 2
C. 4
D. 0
4. D.

For the last digit of $a_{n}\left(n \in \mathbf{N}_{+}\right)$, it can be proven that it cycles with a period of 4 (or 2 or 1), and it only depends on the last digit of $a$. It can be verified that the sum of the last digits of $a^{n}$ for any 10 consecutive $n$ is 7. Therefore, the last digit of the original expression is the last digit of $7 \times 199 + 1 + 6$, which is 0.