Comment by dwlg00

I got c(c(c))(c).

My steps:

  c(g):      (f) => (x) => g(f(x))
  c(c):      (f) => (x) => c(f(x)) = (g) => (y) => f(x)(g(y)), or
             (f) => (x) => (g) => (y) => f(x)(g(y))
  c(c)(c):   (x) => c(c(x)) = (f) => (y) => c(x)(f(y)) = (z) => x(f(y)(z)), or
             (x) => (f) => (y) => (z) => x(f(y)(z)) # close!
  c(c(c)):   (f) => (x) => c(c)(f(x)) = (y) => c(f(x)(y)) = (g) => (z) => f(x)(y)(g(z)), or
             (f) => (x) => (y) => (g) => (z) => f(x)(y)(g(z)) # We just need to substitute f with c
  c(c(c))(c): (h) => c(c)(c(h)) = (g) => c(c(h)(g)) = (f) => (x) => c(h)(g)(f(x)) = h(g(f(x)), or
              (h) => (g) => (f) => (x) => h(g(f(x)))