It is, though that's not very surprising. Here's a quick proof:

S-K combinators[1] are a Turing-complete subset of lambda calculus. Here's an implementation of them:

  k = (x) ->
    (y) -> x
  
  s = (x) ->
    (y) ->
      (z) -> x(z)(y(z))

1: http://en.wikipedia.org/wiki/SKI_combinator_calculus