$n$ and $n^5$ have the same units digit?

We know if unit digits of two numbers are same, their difference is divisible by 10 and vice versa.

Method $1a:$

Using Fermat's Little Theorem $n^5-n\equiv0\pmod 5$

and $n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n-1)(n+1)(n^2+1)$ which is divisible by $n(n-1)$ which is always even

$\implies 2|(n^5-n)$ and we have $5|(n^5-n)$

$\implies n^5-n$ is divisible by lcm $(2,5)=10$

Method $1b:$

As $10=2\cdot5,$

using Fermat's Little Theorem, we have $$n^5-n\equiv0\pmod 5\text{ and } n^2-n\equiv0\pmod 2$$

Now, lcm $(n^5-n,n^2-n)=n(n^4-1,n-1)=n(n^4-1)$ as $(n-1)|(n^4-1)$

$\implies $lcm $(n^5-n,n^2-n)=n^5-n$ which is divisible by $5,2$ hence by lcm$(2,5)=10$

Method $2:$

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

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 $3!=6$

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

Without using any modular arithmetic:

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

$(n-2)(n-1)n(n+1)(n+2)$ is the product of 5 consecutive integers thus divisible by 2 and 5.

$5n(n-1)(n+1)$ is multiple of $5$ and even.