If $5! \cdot 3! = n!$, what is the value of $n$?
This problem involves factorials, which are products of consecutive positive integers starting from 1. For example, $5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120$. To solve for $n$, I need to compare the factors of $5! \cdot 3!$ and $n!$. I can write $5! \cdot 3!$ as $120 \cdot 6 = 720$. Now, I need to find a value of $n$ such that $n! = 720$. I can start by trying some values of $n$ and see if they match. If $n = 6$, then $n! = 6! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 720$, which is exactly what I want. Therefore, $n = 6$ is the correct answer. 