Is an exponential function one-to-one

Answer 1

Yes, if you accept that the definition of the exponential function f is given by the two statements (1) f(0)=1, and (2) f'(x)=f(x) for all real x, then you have from differential equations that the function is represented by a power series $f(x)=sum_{k=0}^\infty \frac{x^k}{k!}$. If you accept that this means that f(x) is everywhere positive, then you have that f is monotone (increasing), which implies that it is one-to-one by the mean value theorem.