How do you simplify \[5!\]?

Hint: The given question is based on the concept of factorials. We will solve this by using the formula of factorial which states; If n is a factorial, that is, \[n!\], then \[n!=n\times (n-1)\times (n-2)\times ...\times 1\], on applying this formula on \[5!\] we get the value of the given factorial as \[120\].

Complete step by step solution:
Factorial of a number can be defined as the product of that number and the numbers, decreasing by one unit each till the sequence reaches 1. That is,
If n is a factorial, that is, \[n!\], then
\[n!=n\times (n-1)\times (n-2)\times ...\times 1\]
According to the given question we have been required to find out the factorial of the given number, so we have \[5!\]
Here, we can compare \[5!\] to \[n!\] and accordingly we will substitute the values in the formula of the factorial of a number.
So, we have \[n=5\]
Proceeding further, we know that factorial is the product of numbers decreasing by 1 unit until it reaches 1.
So, \[n-1=5-1=4\]
We will still decrease further until we reach the value 1, so we have next,
\[n-2=5-2=3\]
And next we have,
\[n-3=5-3=2\]
And now onto the last number we have,
\[n-4=5-4=1\]
We have reached the number 1, so that is, we have all the numbers required in the formula for the factorial of a number. Now we will simply substitute the number in the formula and multiply all the numbers together.
We have,
\[n!=n\times (n-1)\times (n-2)\times (n-3)\times (n-4)\]
\[\Rightarrow 5!=5\times 4\times 3\times 2\times 1\]
\[\Rightarrow 5!=120\]
Therefore, \[5!=120\]

Note:
In case of factorial of any number we have numbers leading up to 1 only and not 0 because if we include 0, then, the product will get 0. And also \[0!=1\], always remember this.