Proof of proposition 1.2:  en+1(x + 1) > en+1(x).

Let's check that this is correct by double induction.  We need only these clauses of the definition of en:

e1(x) = x2 + 2;
en + 2(0) = 2;
en + 2(x + 1) = en + 1(en + 2(x)).
DI 1, 2. e1(x + 1)  = (x + 1)2 + 2 > x2 + 2  = e1(0).
DI 3. Suppose "x, en + 1(x + 1) > en + 1(x).  Then:
a. en + 2(1) = en + 1(en + 2(0)) = en + 1(2)
> 2 = en + 2(0) by proposition 1.1.
b.  Suppose, further, that en + 2(x + 1) > en + 2(x). 
So there exists k > 0 such that 
en + 2(x + 1) = en + 2(x) + k. 
By definition:
en + 2(x + 2) = en + 1(en + 2(x + 1)) = en + 1(en + 2(x) + k) [by the preceding line].
Using the hypothesis, we have that for each i such that k > i,
en + 1(en + 2(x) + i + 1) > en + 1(en + 2(x) + i).
By transitivity,
en + 1(en + 2(x) + k) > en + 1(en + 2(x))
= en + 2(x + 1).