Prove $n^2(n^4-1)$ is divisible by 60 using Mathematical Induction. Base step:
p(2)=4 * 15= 60
Inductive Hypothesis:
Assuming p(k) = $k^2(k^4-1)$ = 60q
Induction:
p(k+1)= $(k+1)^2[(k+1)^4-1]$
      = $(k+1)^2[(k+1)^2 + 1][(k+1)^2 - 1]$
      = $(k+1)^2(k^2+2k+2)(k^2+2k+1- 1)$
      = $(k+1)^2(k^2+2k+2) k (k+2)$
      = $k(k+1)(k+2)(k^2+2k+2)(k+1)$
Now $k(k+1)(k+2)$ is the product of 3 consecutive natural numbers. Their product HAS to be a multiple of 2 and 3 and hence also a multiple of 6. So all that I have to do is prove that $(k^2+2k+2)(k+1)$ is a multiple of 10. But I can't seem to do so. Also through my induction, I have not at all made use of my hypothesis! How do I go about solving this question through the Principle of Mathematical Induction?

Does it have to be by induction?
Write $n^2(n^4-1)= n^2 (n-1) (n+1) (n^2+1) = n(n-1) n (n+1) (n^2+1)$.
$2$ divides $n(n-1)$ and $n(n+1)$ and so $4$ divides $n^2(n^4-1)$.
$3$ divides $(n-1)n(n+1)$ and so $3$ divides $n^2(n^4-1)$.
Since $60=4\cdot3\cdot5$, it remains to prove that $5$ divides $n^2(n^4-1)=n(n^5-n)$. But this follows from Fermat's little theorem and also by induction.
