How does $\exp(x+y) = \exp(x)\exp(y)$ imply $\exp(x) = [\exp(1)]^x$? In Calculus by Spivak (1994), the author states in Chapter 18 p. 341 that 
$$\exp(x+y) = \exp(x)\exp(y)$$ implies 
$$\exp(x) = [\exp(1)]^x$$ 
He refers to the discussion in the beginning of the chapter where we define a function $f(x + y) = f(x)f(y)$; with $f(1) = 10$, it follows that $f(x) = [f(1)]^x$. But I don't get this either. Can anyone please explain this? Many thanks!

I like (as I have done here before) to start with a functional equation
and derive properties of the function.
If $f(x+y) = f(x) f(y)$ and $f$ is differentiable (and non-zero somewhere),
$f(0)  = 1$ and
$f(x+h)-f(x) = f(x)f(h)-f(x)
=f(x)(f(h)-1)
=f(x)(f(h)-f(0))
$
so
$$(f(x+h)-f(x))/h = f(x)(f(h)-f(0))/h.$$
Letting $h \to 0$,
$f'(x) = f'(0) f(x)$.
From this, $f(x) = \exp(f'(0)x)$.
This also works for $\ln$, $\arctan$, and $\sin$ and $\cos$.
Functional equations are fun!
