Consider projective space of dimension n over Fp. How many lines are there?
Note that for each non-zero vector, we get a 'direction'. So there are pn−1 potential directions.
See that for any choice of direction d∈Fp−0, there are (p−1) "linearly equivalent" directions, given by 1⋅d, 2⋅d, \dots, (p−1)⋅d which are all distinct since field multiplication is a group.
Thus, we have (pn−1)/(p−1) lines. This is equal to 1+p+p2+⋯+pn−1, which is p0+p1+⋯+pn−1
If we plug in p=1 (study the "field with one element", we recover ∑i=0n−1pi=n.
Thus, "cardinality of a set of size n" is the "number of lines of n-dimensional projective space over F1!
Since [n]≡{1,2,…,n} is the set of size n, it is only natural that [n]p is defined to be the lines in Fpn. We will abuse notation and conflate [n]p with the cardinality, [n]p≡(pn−1)/(p−1).
Recall that a maximal flag is a sequence of subspaces V1⊆V2⊆⋯⊆V. At each step, the dimension increases by 1, and we start with dimension 1. So we pick a line l1 through the origin for V1. Then we pick a plane through the origin that contains the line l1through the origin. Said differently, we pick a plane p2 spanned by l1,l2. And so on.
How many ways can we pick a line? That's [n]p. Now we need to pick another line orthogonal to the first line. So we build the quotient space Fpn/L, which is Fpn−1. Thus picking another line here is [n−1]p. On multiplying all of these, we get [n]p[n−1]p…[1]p.
In the case of finite sets, this gives us 1⋅2⋅…n=n!.