$30$
divide $n^5n$
for all positive integers $n$
?

 Factorise
$n^22n3$
.  Is
$n^2+n$
always even when$n$
is an integer?  Does
$6$
divide$100002$
? Try to reason about the divisors of both numbers.
 Factorise

How do you split
$30$
into a product of prime factors? 
Can you factorise
$n^5  n$
? 
You should obtain
$n(n1)(n+1)(n^2+1)$
. What can you say about the product of 3 consecutive numbers in terms of divisibility by$2$
and$3$
? 
If you assume that
$5$
divides$n^5n$
, how can you prove that 5 divides$(n+1)^5(n+1)$
?

First notice that
$30=2\cdot3\cdot5$
. Since$2$
,$3$
and$5$
are all coprime we will prove that each of them divides$n^5n$
and hence conclude that their product does too. We now factorise$n^5n$
:$$ \begin{aligned} n^5n &= n(n^41) \\ &= n(n^21)(n^2+1)\\ &= n(n1)(n+1)(n^2+1) \end{aligned} $$
The product$(n1)n(n+1)$
is of 3 consecutive numbers, hence necessarily both 2 and 3 must divide it. We must now prove divisibility by$5$
. There are several approaches to do this, but here we shall use induction. It’s easy to verify the base case for$n=0$
or$n=1$
. Then do the inductive step:$$ \begin{aligned} (n+1)^5  (n+1) &= n^5 + 5n^4 + 10n^3 + 10n^2 + 5n  n \\ &= n^5n + 5(n^4+2n^3+2n^2+n) \end{aligned} $$
First part is divisible by$5$
owing to the induction hypothesis, while the rest is obviously a multiple of 5.