<s>[INST] "Question: Show that $n^5 - n$ is divisible by 30; for all natural n" 
Show that $n^5 - n$ is divisible by $30;$ $\forall n\in \mathbb{N}$

I tried to solve this three-way. And all stopped at some point.
I) By induction:
testing for $0$, $1$ and $2$ It is clearly true.
As a hypothesis, we have $30|n^5-n\Rightarrow n^5-n=30k$.
Therefore, the thesis would $30|(n+1)^5-n-1\Rightarrow (n+1)^5-n-1=30j$. By theorem binomial $(n+1)^5-n-1=n^5+5n^4+10n^3+10n^2+5n+1-n-1$ $=30k+5n^4+10n^3+10n^2+5n.$
It was a little messy and not given to proceed.
II)I tried by Fermat's Little Theorem:
$$30|n^5-n\Rightarrow 5\cdot3\cdot2|n^5-n $$
Analyzing each case, we have; clearly $5|n^5-n$ (Fermat's Little Theorem). Now cases $3|n^5-n$ and $2|n^5-n$ I could not develop.
III)In this I wanted your help especially like to solve using this because this exercise on this handout (Properties of the Greatest Common Divisor). Would show that
$$gcd(n^5-n,30)=30$$
Are detailers, please, because'm coursing Theory of the numbers the first time. In the third semester of the degree course in mathematics
What would be the idea to solve this by using the Greatest Common? This problem in this handout GCD?
 [/INST] Method $1:$
$$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n^2-1)(n^2-4+5)$$
$$=n(n^2-1)(n^2-4)+5n(n^2-1)$$
$$=\underbrace{(n-2)(n-1)n(n+1)(n+2)}_{\text{ product of }5\text{ consecutive integers  }}+5\underbrace{(n-1)n(n+1)}_{\text{ product of }3\text{ consecutive integers  }}$$
Now, we know the product $r$ consecutive integers is divisible by $r!$ where $r$ is a positive integer

Method $2:$
Using Fermat's Little Theorem, $n^p\equiv n\pmod p$ for prime $p$
$\implies 5$ divides $n^5-n$
$n^5-n=n(n^2-1)(n^2+1)=(n^3-n)(n^2+1)$
As $3$ divides $n^3-n\implies 3$ divides $n^5-n$
$n^5-n=n(n-1)(n+1)(n^2+1)=(n^2-n)(n+1)(n^2+1)$
As $2$ divides $n^2-n\implies 2$ divides $n^5-n$
$\implies $lcm $(2,3,5)$ divides $n^5-n$
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