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 :\

Without using the link I gave you (which is overkilling the problem, by the way), you can see that $2^4$ divides $$f(n):=n^7+n^6-n^5-n^4=(n-1)\,n^4\,(n+1)^2$$
by considering the case $n$ is odd and the case $n$ is even.  Since $n-1$, $n$, and $n+1$ are consecutive integers, $3$ must divide $f(n)$.  That is, $2^4\cdot 3=48$ must divide $f(n)$ for all $n\in\mathbb{Z}_{>0}$.  Using Will Jagy's answer, you should be able to deduce that $48$ is indeed the GCD of $f(n)$ for all $n\in\mathbb{Z}_{>0}$.
However, if you want to use the method given in that link, you can show that $f(n)$ is equal to
$$72\cdot 2!\cdot\binom{n}{2}+360\cdot 3!\cdot\binom{n}{3}+404\cdot4!\cdot\binom{n}{4}+154\cdot5!\cdot\binom{x}{5}+22\cdot6!\cdot\binom{n}{6}+7!\cdot\binom{n}{7}\,.$$
Then, the greatest common divisor of the coefficients $72\cdot 2!$, $360\cdot 3!$, $404\cdot 4!$, $154\cdot 5!$, $22\cdot 6!$, and $7!$ is equal to $$\gcd\big(72\cdot 2!,360\cdot3!,404\cdot 4!\big)=48\,.$$
This is the modified content of $f(n)$ as defined in the link above.
P.S.:  I know other people have more or less answered your question, but I am illustrating how to use the more general method given in my link.  You can use this solution for other integer-valued polynomials.
