Question

Recall that the factorial function is defined as n! = 1 · 2 · 3 · ? · (n ? 1) · n and, by convention, 0! = 1. Give a recurrence relation for n!.

Answer 1

Let
`a_n` denote the factorial of any non-negative integer
`n`.

By convention, we take
`0!=1`, so that
`a_0=1`. As
`n!=n\cdot(n-1)\cdot\cdots\cdot2\cdot1` by definition, we can obtain the next integer's factorial,
`a_(n+1)=(n+1)!`, by multiplying the previous term
`a_n=n!` by the next integer.

The recursive definition for the factorial function can then be given by


`\begin{cases}a_0=1\\a_(n+1)=(n+1)a_n&\text{for }n\ge0\end{cases}`