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 :\

The answer is 48.
The polynomial factors to:
((n+1)^2)(n-1)(n^4). Let's look at primes.
So the number must divide n-1, n and n+1. At least one of those will be even, so we can count on 2 dividing it.
However, in more detail, let's look at cases.
If n is even, we can rely on 2 dividing it 4 times due to the n^4 term.
If n is odd, then n+1 and n-1 are both even. However, since every second even number is divisible by 4, one of those two quantities is as well. If n-1 is divisible by 4, n+1 is divisible by 2 and no higher powers of 2, so that leaves the polynomial divisible by 2 at least 4 times. In the other case, n+1 is divisible by 4, so the polynomial is divisible by 2 at least 5 times.
Thus, the polynomial is always divisible by 2 at least 4 times.
The next prime is 3. Exactly one of the three values (n-1, n and n+1) will be a multiple of 3, and sometimes, that number will be n-1, so our mystery divisor will have up to one power of three in it.
For bigger primes than 3, there will always exist values of n for which none of our factored terms will be a multiple of the prime—in general, n=p+2 would not produce a value of the polynomial divisible by p for p>3. Thus, no other primes can divide our mystery divisor, so that leaves it at 16*3=48.
