$$(n+1)^5-(n+1)=(n^5-n)+5n^4+10n^3+10n^2+5n=(n^5-n)+5n\left( n^3+2n^2+2n+1 \right)$$ $$=(n^5-n)+5n(n+1)\left( n^2+n+1 \right)$$ $$=(n^5-n)+5n(n+1)\left( (n-1)^2+3n\right)$$ ...1 $$=(n^5-n)+5n(n+1)(n-1)^2+15n^2(n+1)$$
- is divisible by $30$ by the inductive assumption
- $n(n+1)(n-1)$ is the product of three consecutive integers, thus divisible by $2$ and $3$.
- $n(n+1)$ is even.
This answer has been posted before but there was (n-1)^2-3n instead of + 3n. I have a test in two hours and I just wanted to clear this doubt. Can anyone help? Thanks in advance.