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.

A trick I learnt from Bill Dubuque.

Write it as

$$\frac{n^5 - n}{5} + \frac{n^4 - n}{2} + \frac{n^3 - n}{3} + n$$

(See Bill's answer here: Using congruences, show $\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n$ is integer for every $n$)

One quick way is to recognize that $$\frac{n^5}{5} + \frac{n^4}{2} + \frac{n^3}{3} - \frac{n}{30}$$ is nothing but $1^4 + 2^4 +\cdots + n^4$.

Another way is to look at $6n^5 + 15n^4 + 10n^3 - n$ and prove that $30$ divides it i.e. prove that $2$, $3$ and $5$ divide $6n^5 + 15n^4 + 10n^3 - n$.

Nobody's given the boring but direct brute-force answer.

Several have noted that you want to show that $30$ divides $6 n^5 + 15 n^4 + 10 n^3 - n$, for all $n$. In other words, that $$ 6 n^5 + 15 n^4 + 10 n^3 - n \equiv 0 \pmod {30}$$

There are only 30 possible values of $n$ modulo 30. Just plug them in and see that you get 0 in every case.

(Optional) By factoring 30 into prime powers, we see that to prove it for modulo 30 is the same thing as jointly proving it for modulo 2, 3, and 5. Brute force is easier with these smaller moduli.