Finding the greatest common divisor of $(n+1)(n+2)^2(n+3)^3(n+4)^4$ Determine the largest positive integer that divides $(n+1)(n+2)^2(n+3)^3(n+4)^4$ for all positive integers n.

First, I noticed that out of $n+1, n+2, n+3,$ and $n+4,$ there must be one multiple of $4$, at least one multiple of $3$, and a multiple of $2$ that isn't the multiple of $4$. Since the problem is asking for a minimal case that satisfies these conditions, $n+1$ should be the multiple of $4$, $n+2$ should be the multiple of $3$, and $n+3$ should be the multiple of $2$, thus giving $4 \cdot 3^2 \cdot 2^3 = 288.$
However, I am unsure as to whether or not this answer is correct, as testing cases has consistently yielded higher GCDs than this.

Let $f(x) \in \Bbb Z[x]$. We have $m \mid f(n) \iff f(n)\equiv 0\pmod m$, but when written as this it's clear that $n$ may be replaced by any integer (not only natural numbers) congruent to $n$ modulo $m$, to get a equivalent congruence. The means that
"The largest positive integer that divide $f(n)$ for all positive integers $n$"
is the same as
"The largest positive integer that divide $f(n)$ for all integers $n$"
Now to this particular case:
We have
$$f(-5) =(-5+1)(-5+2)^2(-5+3)^3(-5+4)^4 = (-4)(-3)^2(-2)^3(-1)^4=2^5 3^2$$
(so the answer can't be larger than that) and you argued that $f(n)$ is always multiple of $2^53^2$ so that's the answer.
