If you substitute $n-1$ into the definition, you get
$$(n-1)! = (n-1) \cdot (n-2) \cdot \dotso\cdot2 \cdot 1\;.$$
This is $n!$ except that the initial factor of $n$ is missing, so $n!=n(n-1)!$. In words: Since $n!$ is the product of all numbers up to $n$, it's the product of $n$ with the product of all numbers up to $n-1$.
As to why we can't leave it as it is, we can; I don't see how your book was implying that we can't. The reason it sometimes makes sense to rewrite the factorial like this is mainly that that's useful in inductive proofs, where some property involving a factorial $n!$ is reduced to the corresponding property involving the factorial $(n-1)!$.