Find the value of $6!$

Hint: First, we need to know about the concept of the factorials, which is the function denoted as $n!$ for any given integer $n$ and then its product of the all positive integers are only less than or equals to the given $n$. Like in the decreasing way we need to decreasing the given positive number to the least positive number $1$
In general, we get $n! = n \times (n - 1)! \times ... \times 2 \times 1$

Complete step-by-step solution:
Given that the problem is to find the value of $6!$, that means we have to find the factorials up to the number six. then its product of the all positive integers is only less than or equal to the given $n$ to decrease to the number one.
Hence, we get $6! = 6 \times 5!$
Using the same method of decreasing integer, we get $6! = 6 \times 5 \times 4!$
 Using the same method of decreasing integer, we get $6! = 6 \times 5 \times 4 \times 3!$
Using the same method of decreasing integer, we get $6! = 6 \times 5 \times 4 \times 3 \times 2!$
Using the same method of decreasing integer, we get $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$
Hence using the multiplication operation, we get, $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$.

Note: We mostly use the factorial methods in the concept of the permutation and combination methods
We make use of the multiplication operation to solve the given problem
Hence using simple operations, we solved the given problem.
Note that $0! = 1$