As such, it would indeed be extraordinary to have some non-trivial function transform an irrational number (or two, like e^pi) in an unexpected way into a rational number. I imagine most mathematicians would take a very lopsided bet that they're irrational.
The rest of the reals -- and it is most of that vast sea -- are completely inaccessible, unnameable, forever anonymous.
For things like e^i you almost cannot define the idea of complex powers without knowing the right concepts. The sqrt(2) example clearly converges toward 2 as you stack the tower. Binet's formula was arrived at via derivation from the Fibonacci recursion formula... so anyone who had it would obviously know its significance.
So, to be clear, the only reason why those examples would look strange is if someone showed it to you (armed with only high school mathematics) without explanation. Or if you don't know how to estimate the solution to f^n(1), f(x) = sqrt(2)^x for n, say, in [1,10].
"Legendre's Constant" has none of those features. That said, with today's computational power deriving a sufficiently strong estimate that it "might be 1" isn't so hard. One can imagine that other slower to converge examples exist today
Is it really that simple? sqrt(2)^4 = 4, so the answer should also 4 by the same argument. Which makes the whole expression not well-defined IMHO.
Instead, we can look at it like this: If f is a continuous, order-preserving function from a closed interval to itself, and x <= f(x), then x <= f(x) <= f(f(x)) <= ..., with this sequence converging to the least fixed point >= x.
In our case, we can think of f(x) = sqrt(2)^x as a continuous function from [-infinity, +infinity] to itself, and take our starting point to be x = sqrt(2) [or perhaps 1, or 0, or -infinity, each one step back from the previous]; thus, the sequence sqrt(2), sqrt(2)^sqrt(2), sqrt(2)^(sqrt(2)^sqrt(2)), ..., converges to the least fixed point of f, which will be 2 rather than 4.
1. e^(i * pi) "looks" like nonsense. If you think of exponentiation as iterated multiplication, use algebra (roots) to compute rational exponents, use continuity to fill in the non-integer real values... then, still, how do you "do something" an imaginary number of times? Of course, it's actually -1. This doesn't qualify (IMO) because as soon as mathematicians understood complex exponents, this result was probably fairly obvious.
2. The infinite tower sqrt(2)^sqrt(2)^sqrt(2)^... has a finite value, which is 2. (It's the fixed point, sqrt(2)^x = x). Yet again, we have a simple (integer) value coming out of something that seems infinitely complex. However, yet again, once you make the conceptual leap necessary to evaluate such "towers", the answer is trivial.
3. Binet's formula "looks" like it should be producing irrational numbers for integer inputs, but actually produces the Fibonacci sequence. But this is just high school algebra: the problematic terms cancel and you're left with something reducing to an integer.
I suspect that all of the "long suspected irrationals" like e^pi and pi^e are irrational. Almost all mathematicians, if it were an even-odds bet, would bet on them being so. (Of course, intuition can and does fail in all sorts of ways.) It's just that, for all but the simplest numbers, we don't have the tools to prove irrationality when, in fact, they are. That requires quantifying over all the integers, and we already know that some first-order statements are undecidable. The "scary" (and, in my mind, likely) possibility is that, for some of these simple-to-express real numbers, it's formally undecidable whether they are rational. More weirdly, no specific real number will ever be proven to be in the "unprovably irrational" category (since that would, by definition, constitute proof of irrationality).