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.