Currently I'm reading an awesome book: Probability: For the Enthusiastic Beginner written by David Morin (Harvard university). I highly recommend the book for anyone who like me is a beginner in the field and looks for a beginner friendly book about the subject. In chapter one, where the author discusses combinatorics, he presents how to build ordered sets where repetitions are not allowed, that is,
How many different sets of $n$ objects can be chosen from a given set of $N$ objects, where the order matters and where repetitions are not allowed. we must of course have $n \leq N$, because we can’t use a given object more than once.
The author presented nicely via several examples and figures that the number of ways to build ordered sets of $n$ objects from $N$ objects $(n \leq N)$ is:
$$_NP_n=N(N−1)(N−2)⋅⋅⋅(N−(n−1))$$
So until here, everything is fine and clear. But then the author says
If we multiply this by $1$ in the form of $\frac{(N − n)!}{(N − n)!}$, we see that the number of ordered sets of $n$ objects chosen from $N$ objects can be written in the concise form,
$$_NP_n = \frac{N!}{(N−n)!}$$
In the above formula, in the fraction, I can understand why we have $(N − n)!$ in the denominator but I'm not sure that how the expression ended up as $N!$ in the numerator. It has been a while since I have worked with factorial expressions like this and I'm not sure to follow. Just as an example I tried with $N = 5$ and $n = 2$ just to see how the numerator of the fraction is equal to N! and it seems to me (please correct me if I'm wrong) that the author considers that
$$N! = N(N−1)(N−2)⋅⋅⋅(N−(n−1))\underbrace{(N−n)(N−(n+1))(N−(n+2))...(N−(N−3))(N−(N−2))(N−(N−1))}_\text{So is it correct to say that above expression = $(N-n)!$ ?}$$
And therefore
$$N! = \left(\prod_{k = 0}^{n - 1} (N - k)\right)(N - n)!$$
Am I right?