Perhaps an easier example is to do this in GHCi:

    let x = id id id id id id id id id id id id id id id id id id id id id id id id

See how long that takes (add ids if your machine can take it). To deduce a type for x, ghci will use 2^k type variables, if you have k ids. This is easily seen by induction:

If k = 1, then the type of x is a -> a. This takes 2 = 2^1 = 2^k variables. If k > 1, it is of the form id f. Let t be the type of f. t has 2^{k - 1} variables, by induction. Thus the outermost id must take something of type t (this is 2^{k - 1} variables) and produce something of that same type t (this is another 2^{k - 1} variables), that is, it's of type t -> t. 2^{k - 1} + 2^{k - 1} = 2^k.

Note that this information is purely at compile time. Haskell isn't typed at runtime, this is all just to check types. At runtime, x is a no-op, as expected.