Recursive factorial

For positive values of $n$, let's write $n!$ as we did before, as a product of numbers starting from $n$ and going down to $1: n! = n⋅(n−1)⋯2⋅1$. But notice that $(n−1)⋯2$⋅ $1$ is another way of writing $(n−1)!$, and so we can say that $n! = n⋅(n−1)!$. Did you see what we just did? We wrote $n!$ as a product in ...