Calculate, with proof, the last digit of

$$
3^{3^{3^{3^{3}}}}
$$

Remark. Note that this means $3^{\wedge}\left(3^{\wedge}\left(3^{\wedge}\left(3^{\wedge} 3\right)\right)\right)$, not $\left(\left(\left(3^{\wedge} 3\right)^{\wedge} 3\right)^{\wedge} 3\right)^{\wedge} 3$.
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

$$
n_{4}=3^{3^{3^{3^{3}}}}
$$

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 .