The arrangement formula

Step 1 (set up the positions): label the $k$ positions to be filled, in order, as position 1 through position $k$. Filling each position with one of the $n$ objects, without reusing an object already placed, is a task made of $k$ successive steps.

Step 2 (apply the multiplication rule): position 1 can be filled in $n$ ways. Once it is filled, one object has been used, so position 2 can be filled in $n-1$ ways, regardless of which object filled position 1. Continuing this way, position $k$ can be filled in $n-k+1$ ways, since $k-1$ objects have already been used. By the multiplication rule, the total count is the product $n(n-1)\cdots(n-k+1)$.

Step 3 (rewrite as a factorial quotient): multiplying and dividing by the missing tail $(n-k)!$ gives $n(n-1)\cdots(n-k+1) = \frac{n(n-1)\cdots(n-k+1)\cdot (n-k)!}{(n-k)!} = \frac{n!}{(n-k)!}$, which is exactly $A_n^k = \frac{n!}{(n-k)!}$.

Step 4 (special case $k=n$): substituting $k=n$ gives $A_n^n = \frac{n!}{0!} = \frac{n!}{1} = n!$ using $0! = 1$, which recovers the permutation count $P_n = n!$.