Prove: $\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n {30}$ is an integer for $n \ge 0$ I am attempting to prove the following problem: 

Prove that $\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n {30}$ is an integer for all integers $n = 0,1,2,...$

I attempted to solve it by induction, but when proving for $n= x+1$ the algebra gets very messy very fast. I was wondering if this is the only way or if there is a quicker way to prove this. I guess I am a little unsure as to how to prove something is an integer. 
I also noticed that letting $f(x) = \frac{x^5}5 + \frac{x^4}2 + \frac{x^3}3 - \frac x{30}$ and deriving $f(x)$ yields a fairly clean result, but I don't know if this helps me at all. Any help would be great.

Induction is the way to go, but I'd rather show the following fact:
$\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n {30}$ being an integer is the same as $a_n=6n^5+15n^4+10n^3-n$ being divisible by 30 for all $n\geq 0$. The base case is trivial. Assume that the statement is true for $n$, that is, $6n^5+15n^4+10n^3-n$ is divisible by $30$. Now, we can show that $a_{n+1}-a_n$ is also divisible by $30$ and we will be done. 
$$a_{n+1}-a_n = 6((n+1)^5-n^5)+15((n+1)^4-n^4)+10((n+1)^3-n^3)-((n+1)-n)$$
$$=6(5n^4+10n^3+10n^2+5n+1)+15(4n^3+6n^2+4n+1)+10(3n^2+3n+1)-1$$
$$=30n^4+120n^3+180n^2+120n+30$$
