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 :\
 A: we can use some common sense.
If $n$ is even $16|n^4$.  But if $n$ is odd then $2\not \mid n^4$.  However  $n+1$ and $n-1$ are both even so $2^3|(n-1)(n+1)^2$.  But if $n$ is odd then either $n + 1$ or $n -1$ is divisible by $4$ so $2^4|(n-1)(n+1)^2$.  So $2^4$ will divide all such numbers. 
Can any higher powers of $2$ divide it?  Why should they? Example: if $n=2$ then then number is $16*1*3^2$ which is not divisible by any higher power.
But are there any other factors that must divide it?  Either $n$ or $n-1$ or $n+1$ is divisible by $3$ so $3|n^4(n-1)(n+1)^2$.  But if $3|n-1$ then $1$ is the only guaranteed power of $3$ that divides.  (if $n-1 = 3$ then $9 \not \mid n^4(n-1)(n+1)^2$.
So $3*2^4$ will always be a factor.  
Are there any other prime factors that must occur?  Well, if $p$ is a prime $p > 3$ and if we let $n-1 = p+1$ we have $n^4(n-1)(n+1)^2 = (p+2)^2(p+1)*(p+3)^2$ and $p$ is not a factor of that.
So $48 = 3*2^4$ the largest integer that must divide all $n^4(n-1)(n+1)^2$.
A: I really suggest doing some calculations. By hand. With the four numbers below, we know that the final answer is either $48$ or a divisor of it. I see that Donald has shown that $48$ always divides the polynomial, so that is your answer.
..........................
2  poly  144 =  2^4 3^2   gcd so far  144
3  poly  2592 =  2^5 3^4   gcd so far  144
4  poly  19200 =  2^8 3 5^2   gcd so far  48
5  poly  90000 =  2^4 3^2 5^4   gcd so far  48

========================
I did have the computer factor each number for the purposes of illustration, and such factorization would not be difficult by hand, since the polynomial factors nicely. However, it was not necessary to factor the individual numbers in order to keep a running calculation of the gcd. 
A: We have $3$ consecutive numbers so one of these will give a factor of $3$.
If $n$ is even then $n$ will give a factor of $16$. If $n$ is odd then $(n-1)$ and $(n+1)$ will both be even and if $(n+1)^2$ only gives $2$ factors of $2$ then $(n-1)$ will be divisible by $4$, so $(n-1)(n+1)^2$ will be divisible by $16$.
So $\color{red}{48}$ will always divide $n^7+n^6-n^5-n^4$.
A: 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 $n=4$.
A: 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.
A: 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.
