For \(n\) = 2, this gives 2! = (2)(1) = 2.

For \(n\) = 3, this gives 3! = (3)(2!) = (3)(2)(1) = 6.

Continuing this process we find that for any natural number \(n\),

\[ n! = (n)(n-1)...(2)(1). \]

For example,

\[ 5! = (5)(4)(3)(2)(1) = 120, \]

\[ 7! = (7)(6)(5)(4)(3)(2)(1) \]

\[ = (7)(6)(5!) = 42(120) = 5040. \]

\end{align*}

We extend the factorial function to the natural number \(0\) by the definition:

\[ 0! : = 1. \]

This looks odd but is necessary for the definition of the binomial coefficient to follow.

(The factorial can be defined in terms of the gamma function, which can be defined for any real number, and this gives 0! = 1, as a deductive result, and not a definition.)