Factorial expression of $\frac{n!}{(n-r)!}$ If $n \gt r,$ 
Then how is the following valid?
$$n \cdot (n-1) \cdot (n-2) \cdots (n-(r-1))=\frac{n!}{(n-r)!}?$$
I thought it the list of multiplications would be equal to $\frac{n!}{(n-(r-1))!}?$

$\require{cancel}$Maybe a concrete example will make things clearer:
\begin{align}
& \frac{9!}{(9-4)!} = \frac{9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{5\cdot4\cdot3\cdot2\cdot1} = \frac{9\cdot8\cdot7\cdot6\cdot\cancel{5\cdot4\cdot3\cdot2\cdot1}} {\cancel{5\cdot4\cdot3\cdot2\cdot1}} = 9\cdot8\cdot7\cdot6 \\[12pt]
= {} & \underbrace{9\cdot8\cdot7\cdot 6}_{\text{4 factors since} \\  9 \,-\, 4 \text{ rather} \\ \text{than } 9 \text{ minus} \\ \text{something else} \\
\text{appeared.}} \quad \text{and the last factor, 6, is } (9-4+1).
\end{align}
