Can I prove there is no real solution except $x=0, x=1$, without using the function $W(x)$?
And is it possible to do it without using calculus?
Here is my attempts:
$2^x>0 \Rightarrow x+1>0 \Rightarrow x>-1$.
Now, I need to look at these intervals.
$$x\in (-1,0]; [0,1]; [1, \infty)$$
Maybe, it is easy to prove there is no real solution for $x>1$. Because , for $x\to\infty$, we get $2^x>>x$.
Problematic point is , $x\in [0,1]$ or $x\in [-1,0]$.
For $0<x<1$, we get $2>2^x>1$, which is correct ,because $2>x+1>1$ also true. So, this method doesn't work. I need more rigorous method.