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...
 A: An alternate technique (not using induction): $n^5-n=n(n^2+1)(n+1)(n-1)$. Clearly $6\mid (n-1)n(n+1)$, and $5\mid n^2+1$ if $n\equiv 2,3\pmod{5}$, otherwise $n\equiv 0,1,4\pmod{5}$, in which case $5\mid n,n-1,n+1$, respectively.
A: 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 $3$.
A: 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 
$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n-1)(n+1)(n^2+1)=(n^2-n)(n+1)(n^2+1)$ 
A: Using induction.
$n=1$: $1^5-1=0$ is divisible by $30$.
$(n+1)^5 - (n+1) = n^5 - 5n^4 + 10n^3 - 10n^2 + 5n - 1 - n + 1$.
Remember that the assumption is that $n^5 - n$ is divisible by $30$. Subtract the assumption from the above and so we need to prove that   $- 5n^4 + 10n^3 - 10n^2 + 5n$ is divisible by $30$.
To show divisibility by $30$ you can show divisibility by all of $5, 3, 2$.
The above is clearly divisible by $5$ as it's $5 \times(-n^4 +2n^3 \ldots$
After dividing by $5$ we can also represent the above as $-n\times(n^3 -2n^2 +2n -1)$.
This is divisible by $2$, because either $n$ is even or if it is odd then $(n^3 -2n^2 +2n -1)$ is even.
Now we need to prove that $-n\times(n^3 -2n^2 +2n -1)$ is divisible by $3$.
$n$ is of the form of either $3N+0$ or $3N+1$ or $3N+2$.
With some labor you can show that in all cases the above is divisible by $3$.
A: Let $f(m)=m^5-m$
$\displaystyle\implies f(1)=1^5-1=0$
Let $f(m)$ is divisible by $30$ for $m=n$
Now $\displaystyle f(n+1)-f(n)=(n+1)^5-(n+1)-(n^5-n)=5n^4+10n^3+10n^2+5n$
$\displaystyle=5n(n^3-n)+10(n^3-n)+15n^2+15n$
$\displaystyle=5(n+2)n(n+1)(n-1)+15n(n+1)$
Observe that $n(n+1)(n-1),$ being product of $3$ consecutive integers is divisible by $3$ 
and $n(n+1)$ is divisible by $2$
$\displaystyle\implies n(n+1)(n-1)$ is divisible by lcm$(2,3)=6$
$\displaystyle\implies f(n+1)-f(n)$ is divisible by $30$
But, $f(n)$ is divisible by $30,\implies f(n+1)$ will be divisible by $30$
