The last digit of $2^{2006}$ My $13$ year old son was asked this question in a maths challenge. He correctly guessed $4$ on the assumption that the answer was likely to be the last digit of $2^6$. However is there a better explanation I can give him?

The last digit of the number is the remainder of division by 10, since any number is represented as:
$$ a_n \cdot 10^n + a_{n-1} \cdot 10^{n-1} + \ldots + a_1 \cdot 10 + a_0 = \overline{a_na_{n-1}\ldots a_1a_0} $$ $$\quad \Leftrightarrow \quad $$ $$a_n \cdot 10^n + a_{n-1} \cdot 10^{n-1} + \ldots + a_1 \cdot 10 + a_0 \equiv a_0 \mod 10. $$
Therefore, it is enough to consider the remainder of division by $10$ of the original number.
According to the Chinese remainder theorem
$$ 2^{2006} \equiv x \mod 10 \quad \Leftrightarrow \quad \begin{cases} 2^{2006} \equiv x \mod 5 \\ 2^{2006} \equiv x \mod 2 \end{cases} \quad \Leftrightarrow \quad \begin{cases} 2^{2006} \equiv x \mod 5 \\ 0 \equiv x \mod 2\end{cases}$$
Since the Euler function of 5 is equal to $\phi(5) = 5-1 = 4$ (as well as for any prime number), then by Euler's theorem the system of comparisons can be rewritten as:
$$ \begin{cases} 2^2 \equiv x \equiv 4 \mod 5 \\ 0 \equiv x \mod 2\end{cases}$$
Going through all the numbers from $0$ to $10$, we make sure that only $4$ is suitable, so $x = 4$, which will be the last digit of the number.
