1. Since $47 \equiv 7(\bmod 10), 47^{2} \equiv 7^{2} \equiv 49 \equiv-1(\bmod 10)$, $47^{4} \equiv(-1)^{2} \cong 1(\bmod 10)$.
(*)
Now consider $47^{47^{\cdots 47}}$ ( $k$ 47s) with the exponent $47^{\cdots 47}(k-1$ 47s) modulo 4.
Since $47 \equiv-1(\bmod 4)$, we have
$$
47^{47^{\cdots} 47}(k-1 \text { 47s}) \equiv-1 \equiv 3(\bmod 4) \text {. }
$$

Thus, by (*) we get
$$
47^{17^{\cdots 47}}(k \text { 47s}) \equiv 47^{3} \equiv 7^{3} \equiv-7 \equiv 3(\bmod 10) \text {. }
$$

Therefore, the required unit digit is 3.
1. Find the unit digit of $47^{47^{174}}$, where there are $k(>1)$ 47s.