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(define (double f) (lambda (x) (f (f x)))) (( (double (double double) ) inc) 5)
I've got this explanation to fully understand what is happening behind the scene
; explicacion: double1(f) = f (f) double2 double1 = double1 (double1) double3 double2 = double2 (double2) = double1 (double1) (double1 (double1)) double1(inc) = inc(inc) double1 (double1) (double1 double1(inc)) double1 (double1) (double1 (inc(inc))) double1 (f (f)) (inc(inc(inc(inc)))) f(f(f(f))) (inc(inc(inc(inc)))) 16 times inc ; en español: 16 veces inc => 16 times inc(5) = 21
if we have n doubles, there will be 2^{2^(n-1)} times inc, not 2^n times inc.