Showing "$30$ divides $n^5-n$ for all $n\in\Bbb N$" using induction 
Prove that $(n^5 - n)$ divides by $30$ for every $ n\in N$, using induction only. 

How on earth do I do that? Thing is  $(n^5 - n)$  can't be opened using any known formula...

Without using induction
$$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n^2-1)(n^2-4+5)$$
$$=\underbrace{n(n+1)(n-1)(n+2)(n-2)}_{\text{product of }5\text{ consecutive integers }}+5\underbrace{n(n+1)(n-1)}_{\text{product of }3\text{ consecutive integers }}$$
Now from this or this, the product of $n$ consecutive integers is divisible by $n!$
Alternatively, using Fermat's Little theorem, 
$n^5-n$ is divisible by $5$
$n^3-n$ is divisible by $3$
$n^2-n$ is divisible by $2$
Now, $n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=(n^2+1)(n^3-n)$ 
and 
$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n-1)(n+1)(n^2+1)=(n^2-n)(n+1)(n^2+1)$ 
