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.

Following the comment of lulu, look at numbers of the form $n=19+24k$.
For $k=0$, $$(n+1)(n+2)^2(n+3)^3(n+4)^4=20\cdot 21^2\cdot 22^3\cdot 23^4$$ which has the factorization $2^5\cdot 3^2\cdot 5\cdot 7^2\cdot 11^3\cdot 23^4$ and is not divisible by any number containing more than five factors of $2$ and two factors of $3$
For $k=1$, the product $44\cdot 45^2 \cdot 46^3\cdot 47^4$ has no factors of $7$, so $7$ cannot be a factor of the divisor you seek.
For $k=2$, the product $68\cdot 69^2 \cdot 70^3\cdot 71^4$ has no factors of $11$, so $11$ cannot be a factor of the divisor you seek.
For $k=4$, the product $116\cdot 117^2 \cdot 118^3\cdot 119^4$ has no factors of $5$ or $23$, so neither $5$ nor $23$ can be a factor of the divisor you seek.
So the only factor of $20\cdot 21^2\cdot 22^3\cdot 23^4$ that can and must appear in the divisor you seek is $2^53^2$.
It's hard to see that in comparing two examples at a time, as many of them will have a $5$ or a $7$, as well as the occasional larger prime factor in common.
