6. The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=3, a_{n}=3^{a_{n-1}}(n \geqslant 2)$, then the last digit of $a_{1990}$ is
A. 3
B. 9
C. 7
D. 1
6. Choose C.

Slightly solved: When $n=1$, $a_{1}=4 \times 10+3$; when $n=2$, $a_{2}=4 \times 6+3$. By mathematical induction and the binomial theorem, it can be proven that $a_{n}=4 m+3(m \in \mathbf{N})$. Therefore, $a_{1990}=3^{a_{1990}}=3^{4 m+3}=81^{m} \cdot 27$. Hence, the last digit of $a_{1990}$ is 7.