elementary number theory - 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!)

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

Answer

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 $n^4(n-1)(n+1)^2$.