Find the maximum positive integer that divides $n^7+n^6-n^5-n^4$ 
Find the maximum positive integer that divides all the numbers of the form $$n^7+n^6-n^5-n^4 \ \ \ \mbox{with} \ n\in\mathbb{N}-\left\{0\right\}.$$

My attempt
I can factor the polynomial 
$n^7+n^6-n^5-n^4=n^4(n-1)(n+1)^2\ \ \ \forall n\in\mathbb{N}.$
If $n$ is even then exists $\,k\in\mathbb{N}-\left\{0\right\}\,$ such that $n=2k.\;$ so: 
$n^4(n-1)(n+1)^2=2^4 k^4(2k-1)(1+2k)^2$
If $n$ is odd then exists $k\in\mathbb{N}$ such that $n=2k+1$ so
$$n^4(n-1)(n+1)^3=2^3(2k+1)^4(k+1)^2$$
Can I conclude that the maximum positive integer that divides all these numbers is $N=2^3?$ (Please, help me to improve my english too, thanks!)

Note: I correct my "solution" after a correction... I made a mistake :\

Newton's interpolation formula
gives
$$
n^7+n^6-n^5-n^4
=144 \binom{n}{2}+ 2160 \binom{n}{3}+ 9696 \binom{n}{4}+ 18480 \binom{n}{5}+ 15840 \binom{n}{6}+ 5040 \binom{n}{7}
$$
The gcd of the coefficients in the binomial representation is $48$ and so  $n^7+n^6-n^5-n^4$ is always a multiple of $48$.
$48$ is largest factor of all values of $n^7+n^6-n^5-n^4$ because it is the gcd of the values for $n=2$ and $n=4$.
