Four. (20 points) The sequence $\left\{a_{n}\right\}$ is defined as follows: $a_{1}=3, a_{n}=$ $3^{a_{n-1}}(n \geqslant 2)$. Find the last digit of $a_{n}(n \geqslant 2)$.
Let's prove, when $n \geqslant 2$, we have
$$
a_{n}=4 m+3, m \in \mathbf{N} \text {. }
$$

Using mathematical induction:
(i) When $n=2$, $a_{2}=3^{3}=4 \times 6+3$, so equation (1) holds.
(ii) Assume when $n=k(k \geqslant 2)$, equation (1) holds, i.e.,
$$
a_{k}=4 m+3, m \in \mathbf{N} \text {. }
$$

Then $a_{k+1}=3^{a_{k}}=3^{4 m+3}=(4-1)^{4 m+3}$
$$
\begin{aligned}
= & \mathrm{C}_{4 m+3}^{0} \cdot 4^{4 m+3}(-1)^{0}+\mathrm{C}_{4 m+3}^{1} 4^{4 m+2}(-1)^{1} \\
& +\cdots+\mathrm{C}_{4 m+3}^{4 m+2} 4^{1}(-1)^{4 m+2} \\
& +\mathrm{C}_{4 m+3}^{4 m+3} 4^{0}(-1)^{4 m+3} \\
= & 4 t-1=4(t-1)+3 .
\end{aligned}
$$

Here, $t=\mathrm{C}_{4 m+3}^{0} 4^{4 m+2}+\mathrm{C}_{4 m+3}^{1} 4^{4 m+1}(-1)^{1}+\cdots+$ $\mathrm{C}_{4 m+3}^{4 m+2}(-1)^{4 m+2}$ is an integer, and since $\left\{a_{n}\right\}$ is an increasing sequence, we have $a_{k+1} \geqslant a_{k} \geqslant \cdots \geqslant a_{2}=27$, i.e., $4 t-1 \geqslant 27$. Thus, $t \geqslant 7, t-1 \geqslant 6$, i.e., $t-1 \in \mathbf{N}$.
Therefore, when $n=k+1$, equation (1) holds.
Thus, equation (1) holds, and we have
$$
a_{n}=3^{a_{n-1}}=3^{4 m+3}=(81)^{m} \cdot 27 \text {, }
$$

and its last digit is 7.