Why is the last digit of $n^5$ equal to the last digit of $n$? I was wondering why the last digit of $n^5$ is that of $n$? What's the proof and logic behind the statement? I have no idea where to start. Can someone please provide a simple proof or some general ideas about how I can figure out the proof myself? Thanks.

If $\gcd(a, n) = 1$ then by Euler's theorem,
$$a^{\varphi(n)} \equiv 1 \pmod{n}$$
From the tables and as @SeanEberhard stated, $$ \varphi(10) = \varphi(5*2) = 10\left( 1 - \frac{1}{5} \right) \cdot \left(1 - \frac{1}{2} \right)$$
$$= 10\left(\frac{4}{5} \right) \cdot \left(\frac{1}{2} \right) = 4$$
Let $n=10$ and thus,
$$a^{\varphi(10)} \equiv 1 \pmod{10} \implies a^{4} \equiv 1 \pmod{10}$$
Multiply both sides by $a$,
$$a^{5} \equiv a \pmod{10}$$
