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...

If you do want to stick with induction:
The base for $n=0$ (or $n=1$) is clear.
Assume that $n^5-n$ is divisible by $30$. You need to prove that $(n+1)^5-(n+1)$ is also divisible by $30$. Expanding, we get:
$$\begin{align*}
(n+1)^5-(n+1) &= n^5+5n^4+10n^3+10n^2+5n+1-n-1\\
 &= (n^5-n)+5(n^4+2n^3+2n^2+n)\\
 &= (n^5-n)+5n((n+1)^3-n(n+1))\\
 &=(n^5-n)+5n(n+1)(n^2+n+1)
\end{align*}$$
By IH, $n^5-n$ is divisible by $30$, so it is left to show that $n(n+1)(n^2+n+1)$ is divisible by $6$. It easy to see that it is divisible by $2$. Show that it is divisible by $3$.
