Berkeley Math Circle Monthly Contest 5

Solution:
When $3$ is raised to the successive powers $1, 2, 3, 4, \ldots$, the units digits are $3, 9, 7, 1, \ldots$ From then on, since the digit $1$ has been reached, the units digits will repeat in this cycle of four elements. So it is necessary to find the remainder when
$$n_{4} = 3^{3^{3^{3}}}$$
is divided by $4$.
When powers of $3$ are divided by $4$, the remainders are $3, 1, \ldots$ Here the number $1$ appears after two steps, and the remainders therefore repeat in a two-element cycle. So it is necessary to find the remainder when
$$n_{3} = 3^{3^{3}}$$
is divided by $2$.
But $n_{3}$ is clearly odd, so $n_{4}$ has a remainder of $3$ when divided by $4$ and the original number $n_{5}$ has a last digit of $7$.