$n$ and $n^5$ have the same units digit? Studying GCD, I got a question that begs to show that $n$ and $n^5$ has the same units digit ...
What would be an idea to be able to initiate such a statement?
testing
$0$ and $0^5=0$
$1$ and $1^5=1$
$2$ and $2^5=32$
In my studies, I have not got "mod", please use other means, if possible of course.
I demonstrated in a previous period that
$$2|n^5-n$$because $$n^5-n=(n+1)n(5n^4+5n+5)$$, 
and$$5|n^5-n$$By Fermat's Little Theorem
Only I do not understand what should happen to the units of the two numbers are equal ... What must occur?

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 $3!=6$
$$\implies n^5-n\equiv0\pmod{30}\equiv0\pmod{10}$$
