Why is the last digit of $n^5$ equal to the last digit of $n$?

Alternatively, you could prove that $n^5-n$ is divisible by $10$ by induction. If $n=0$, it is obviously true.

Assume true for $n$, we need to show that:

$$(n+1)^5-(n+1) = n^5 + 5n^4 + 10n^3 + 10n^2 + 5n + 1 - (n+1) \\=n^5 - n + 10(n^3+n^2) +5n(n^3+1)$$

is divisible by 10.

But $n^5-n$ is divisibly by $10$ by induction, and $10(n^3+n^2)$ is obviously divisible by $10$, so all you need to show is that $5n(n^3+1)$ is divisible by $10$, which is the same as proving that $n(n^3+1)$ is divisible by $2$.

The fundamental reason for this, as everybody has noted, is due to rules of modular arithmetic.

Note, the last digit of $n^5$ can only be affected by the last digit of $n$. That is, the 1s digit of any power of $n$ will only be affected by the ones digit of $n$, and not the 10s, 100s, or any other digit. So, you only have 10 cases, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. All you have to do is try them separately. $0^5 = 0$, $1^5 = 1$, $2^5 = 32$, $3^5 = 243$, ..., $9^5 = 59049$.

If you've heard of modular arithmetic, all numbers mod 10 are going to be in the set $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$. And, this is why we only need consider these 10 cases.

$$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n^2-1)(n^2-4+5)$$ $$=n(n^2-1)(n^2-4)+5n(n^2-1)$$ $$=\underbrace{(n-2)(n-1)n(n+1)(n+2)}_{\text{ product of }5\text{ consecutive integers }}+5\cdot \underbrace{(n-1)n(n+1)}_{\text{ product of }3\text{ consecutive integers }}$$

Now, we know the product $r$ consecutive integers is divisible by $r!$ where $r$ is a positive integer

So, $(n-2)(n-1)n(n+1)(n+2)$ is divisible by $5!=120$ and $(n-1)n(n+1)$ is divisible by $3!=6$

$$\implies n^5-n\equiv0\pmod{30}\equiv0\pmod{10}$$