Why is it that the Lambert W relation cannot be expressed in terms of elementary functions? According to this Wikipedia page, the Lambert W relation cannot be expressed in terms of elementary functions.
However, it does not explain why this is the case.
An elementary function is "a function of one variable which is the composition of a finite number of arithmetic operations (+ – × ÷), exponentials, logarithms, constants, and solutions of algebraic equations (a generalization of nth roots)," according to Wikipedia. 
Note
I know that I should not always refer to the same source for information, but I believe that this is an accurate definition of the term.
Questions
Does there exist a proof that the Lambert W relation cannot be expressed in terms of elementary functions?
Why is it that the Lambert W relation cannot be expressed in terms of elementary functions?
 A: The branches of Lambert W are the local inverses of the Elementary function $f$ with $f(z)=ze^z$, $z \in \mathbb{C}$.
The incomprehensibly unfortunately hardly noticed theorem of Joseph Fels Ritt in Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90 answers which kinds of Elementary functions can have an inverse which is an Elementary function. It is also proved in Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J. Math 101 (1979) (4) 743-759.
Ritt, J. F.: Integration in finite terms. Liouville's theory of elementary methods. Columbia University Press, New York, 1948
The non-elementarity of LambertW was already proved by Liouville in
Liouville, J.: Mémoire sur la classification des transcendantes et sur l'impossibilité d'exprimer les racines de certaines équations en fonction finie explicite des coefficients. Journal de mathématiques pures et appliquées 2 (1837) 56–105, 3 (1838) 5233–547
It is also proved in
Rosenlicht, M.: On the explicit solvability of certain transcendental equations. Publications mathématiques de l'IHÉS 36 (1969) 15-22
and in
Bronstein, M.; Corless, R. M.; Davenport, J. H., Jeffrey, D. J.: Algebraic properties of the Lambert W Function from a result of Rosenlicht and of Liouville. Integral Transforms and Special Functions 19 (2008) (10) 709-712.
Ritt's theorem shows that no antiderivatives, no differentiation and no differential fields are needed for defining the Elementary functions.
A: I would recommend that you could possibly look into hyper-operations.  They are quite interesting and related.  For quick reference, a hyper-operation of n is a repetition of they {n-1}th hyper-operation.$$5*4=5+5+5+5$$Here, multiplication was turned into a repetition of addition.  Similarly, exponentiation is a repetition of multiplication.  And so on...
But more importantly, all of this can be represented in terms of elementary functions.  Whether it be addition, multiplication, or exponentiation.
Furthermore, the inverses of each individual hyper-operation is considered elementary.  The opposites of addition, multiplication, and exponentiation are subtraction, division, and logarithms or roots, respectively.
However, a combination of different levels of hyper-operations (except addition) cannot be inverted.  For example:$$f(x)=xe^x$$$$f^{-1}=?$$For this example, we assign the Lambert W function as the solution.$$f^{-1}=W(x)$$But the problem is that we cannot turn this into something involving only addition, multiplication, exponentiation, and their inverses (or higher hyper-operations like tetrations and such).
More specifically, we cannot turn this into a $finite$ amount of terms being added or multiplied and such.
To answer your first question, the proof is simply that the definition of the Lambert W function cannot be solved with elementary functions.
As to why, it is because, as far as I can explain, two hyper-operations, multiplication and exponentiation, were combined.  In general, combining different hyper-operations results in unsolvable inverses (or a manipulation of the Lambert W function).  Try solving (without the Lambert W function): $$f(x)=x^a+bx$$$$f^{-1}(x)=?$$Now try solving it with $a=3,2,1,0$.  Much easier?  To solve for any known $a$ is easy, but solving for all $a$'s is more difficult.
The reason why you were able to solve for the above values of $a$ were most likely because of factoring.  However, you cannot factor with exponents like you can with polynomials.  Envision the following:$$a^{(ax)^{(ax^2)^{..^{..^{..}}}}}$$Compare it to$$a+ax+ax^2+...$$$$and$$$$a*ax*ax^2*ax^3*...$$The last two are simplifiable, but the exponential one was not.  This is why we stop being able to find inverses of functions when they are exponential, tetrational, or higher.
