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.

First write $f(n)=(n+1)(n+1)^2(n+2)^3(n+4)^4$ The gcd of the $f(n)$s; $n \in \mathbb{N}$, is of the form $2^b3^a$, for some positive integers $a$ and $b$. Indeed, no prime $p \ge 7$ divides $f(1)$, and $5$ does not divide $f(5)$. Thus indeed, the  common prime divisors are restricted to $2$ and $3$, and thus the gcd of the $f(n)$s is indeed of the form $2^b3^a$.
We now find $b$. Observe that the largest power of $2$ that divides $f(3)$ is $5$. Meanwhile, the largest power of $3$ that divides $f(1)$ is $2$. So $a \le 2$ and $b \le 5$. As you have shown $a \ge 2$ and $b \ge 5$, it follows that the equations $a=2$ and $b=5$ hold. This leaves us with the gcd of the $f(n)$s; $n \in \mathbb{N}$, is indeed $2^53^2$.
