Prove that $n^{30}-n^{14}-n^{18}+n^2$ is divisible by $46410$ I have the following question.

Prove that $n^{30}-n^{14}-n^{18}+n^2$ is divisible by $46410$ for all positive integer $n$.

My attempt:
Firstly, notice that $$n^{30}-n^{14}-n^{18}+n^2$$
$$=n^2(n-1)^2(n+1)^2(n^2+n+1)(n^2-n+1)(n^2+1)^2(n^4-n^2+1)(n^4+1)(n^8+1).$$
Also, $$46410=2\times3\times5\times7\times13\times17$$
My first thought was to use induction, maybe substitute $n+1$ into $n$, then find terms that was already divisible by $46410$, but then I don't think this is a smart way to do it.
Are there any better ways to do it?

Let $$f(n):=n^{30}-n^{18}-n^{14}+n^2\,.$$
Thus,
$$f(n)=\left(n^{13}-n\right)\,\left(n^{17}-n\right)\,.$$
Set $r(n):=n^{13}-n$ and $s(n):=n^{17}-n$.
From here, you can see that
$$D(13,1)=\{3,5,7,13\}\,.$$
Thus,
$$g(13,1)=2\cdot 3\cdot 5\cdot 7\cdot 13\,.$$
Therefore, $2\cdot 3\cdot 5\cdot 7\cdot 13$ is a factor of $r(n)$, whence also a factor of $f(n)$, for every integer $n$.
On the other hand, the same link gives
$$D(17,1)=\{3,5,17\}\,.$$
Consequently,
$$g(17,1)=2\cdot 3\cdot 5\cdot 17\,.$$
Ergo,
$2\cdot 3\cdot 5\cdot 17$
is a factor of $s(n)$, whence also a factor of $f(n)$, for each $n\in\mathbb{Z}$.
As a result, 
$$(2\cdot 3\cdot 5\cdot 7\cdot 13)\cdot (2\cdot 3\cdot 5\cdot 17)=2^2\cdot 3^2\cdot 5^2\cdot 7\cdot 13\cdot 17=1392300=30\cdot 46410$$ 
is a factor of $f(n)$ for every $n\in\mathbb{Z}$.   In fact, 
$$\underset{n\in\mathbb{Z}}{\gcd}\,f(n)=\gcd\big(f(2),f(3)\big)=2^2\cdot 3^2\cdot 5^2\cdot 7\cdot 13\cdot 17=1392300\,.$$
