What I don't get is when it decrements and checks if we're at the base case?
succ = \n\f\x. n f (f x) -- the successor function for Church numerals
F = \c\n. n (c (succ n))
fac = \n\f. n F (\x.f) id
As you can see there is no use of a decrement function. Instead, numbers 1..n are recreated by the iterated successor function.We can follow the reduction of fac 3 as
0 = \f\x. x
1 = \f\x. f x
2 = \f\x. f(f x)
3 = \f\x. f(f(f x))
fac 3 = \f. F (F (F (\x.f))) 1
= \f. 1 (F (F (\x.f)) 2)
= \f. 1 (2 (F (\x.f) 3))
= \f. 1 (2 (3 ((\x.f) 4)))
= \f. 1 (2 (3 f))