$n$ 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?
 A: 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.
A: The result can be verified using minimal machinery.  first note that the units digit of $n^5$ is completely determined by the units digit of $n$. 
The units digit of $n^5$ is one of $0,1,2,\dots,9$. We can verify the result for each of the $10$ cases by a direct calculation. 
There are $10$ calculations to do, none of them painful.  We do one of them. Let $n$ end in $8$. Then $n^2$ ends in $4$, so $n^4$ ends in $6$. Thus $n^5$ ends in $8$. 
A: You can write $n=10a+b$, with $b$ the units digit.  Then $n^5=(10a)^5+5(10a)^4b+10(10a)^3b^2+10(10a)^2b^3+5(10a)b^4+b^5$.  As all the terms but the last have a factor of $10$, the units digit of $n^5$ is the same as the units digit of $b^5$.  This justifies testing each units digit to see if the units digit of its fifth power is the same as the digit.  Ten tests and you are done-they all succeed.
A: 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 $3!=6$
$$\implies n^5-n\equiv0\pmod{30}\equiv0\pmod{10}$$
