46 Sequence $\left\{a_{n}\right\}: a_{1}=3, a_{n}=3^{a_{n-1}}(n \geqslant 2), a_{1990}$ The last digit of is
A. 3
B. 9
C. 7
D. 1
46 C. When $n=1$, $a_{1}=4 \times 0+3$; when $n=2$, $a_{2}=4 \times 6+3$. By mathematical induction and the binomial theorem, it can be proven that $a_{n}=4 m+3, m \in \mathbf{N} \cup\{0\}$. Therefore, $a_{1990}=3^{a_{1} 9_{89}}=3^{4 m+3}=\left(3^{4}\right)^{m} - 3^{3}=(81)^{m} \cdot 27$, where $m$ is a non-negative integer. Thus, the last digit of $a_{1990}$ is 7.