Prove that $8640$ divides $n^9 - 6n^7 + 9n^5 - 4n^3$.
$$ \begin{align} &n^9-6n^7+9n^5-4n^3\\ &\small=362880\binom{n}{9}+1451520\binom{n}{8}+2298240\binom{n}{7}+1814400\binom{n}{6}\\ &\small+734400\binom{n}{5}+138240\binom{n}{4}+8640\binom{n}{3}\\ &=\small8640\left[42\binom{n}{9}+168\binom{n}{8}+266\binom{n}{7}+210\binom{n}{6}+85\binom{n}{5}+16\binom{n}{4}+\binom{n}{3}\right] \end{align} $$
We can further note that $-1$ is a root of $n^3-3n-2$, and that $1$ is a root of $n^3-3n+2$.
You can find these roots by the Rational Root theorem:
Rational root theorem. All rational roots have the form $\frac{p}{q}$,with $p$ a divisior of the constant term and $q$ a divisior of the first coefficient.
Other techniques can also be found in this question and its answers.
Then we simplify it to \begin{align*} n^3(n+1)(n^2-n-2)(n-1)(n^2+n-2) &= n^3(n+1)(n-2)(n+1)(n-1)(n+2)(n-1)\\ &=n^3(n+1)^2(n-1)^2(n-2)(n+2) \end{align*}
There are 6 factors two in 8640.
- If $n$ is even, $n, n-2$ and $n+2$ are even, and furthermore at least one of them is divisible by $4$, which gives 6 factors 2.
- If $n$ is odd, $n-1$ and $n+1$ are even, and furthermore at least one of them is divisible by $4$, which gives 6 factors 2.
There are 3 factors three in 8640.
- If $n \equiv 0 \mod 3$, then $n$ is divisible by 3, and then $27 \mid n^3$.
- If $n \equiv 1 \mod 3$, then $n-1$ and $n+2$ are divisible by 3.
- If $n \equiv 2 \mod 3$, then $n+1$ and $n-2$ are divisible by 3.
The factor 5 can be taken care of since the product is a multiple of five consecutive integers, namely a product of $(n-2)(n-1)n(n+1)(n+2)$.
If you do complete factorisation you get $$n^9−6n^7+9n^5−4n^3=n^3(n-2)(n-1)^2(n+1)^2(n+2)$$ and $$8640=2^6\cdot 3^3\cdot 5$$ Now just look at different cases. For $2\mid n$ we have $2^3\mid n^3$, $2\mid n-2$, $2\mid n+2$ and either $n-2$ or $n+2$ has the factor $2^2$, so we get $2^6\mid n^9−6n^7+9n^5−4n^3$. Do it for the case $n$ odd and for the other factors in a similar style.