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!

As Paul Pichaureau has mentioned, it is easy to see that the formula holds for $n$ natural. You can also easily go further: what should $\exp(-n)$ be? Well, $\exp(0) = 1 = \exp(-n+n) = \exp(-n) \exp(n)$, so $\exp(-n) = 1/\exp(n)$. Furthermore, it is easily seen that $\exp(nx) = \exp(x)^{n}$, so that $$\exp(1) = \exp(n\cdot 1/n) = \exp(1/n)^n,$$ and hence $$\exp(1/n) = \exp(1)^{1/n}$$ by the usual rules for powers. Combining these, we see that $$\exp(p/q) = \exp(1)^{p/q},$$ so the formula holds for any rational number $x$. I think this is about as far as you can get using only that formula (in addition to the requirement that $\exp(0) = 1$). If we, in addition, assume that $\exp$ is continuous on $\mathbb{R}$, then we can choose a sequence $(r_{i})_{i =1}^{\infty}$ of rational numbers such that $\lim_{n \to \infty} r_{i} = x$, and compute 
$$ \lim_{n\to \infty} \exp(1)^{r_n} = \lim_{n \to \infty} \exp(r_{n}) = \exp(\lim_{n \to \infty} r_{n}) = \exp(x).$$ The problem now is what $\exp(1)^x$ is supposed to be. If we already know that the power function $x \mapsto \exp(1)^x$ is continuous, then certainly the right hand side is equal to $\exp(1)^x$, but as Paul mentions, the power functions are often defined in terms of the exponential function. However, I don't think you have to do it that way (you could define rational powers as usual, and prove that a continuous extension to the reals exists), so it all depends on what Spivak has done prior to that chapter (you also didn't specify what $x$ is, so it conceivable that he assumes it is rational, in which case it doesn't matter much).   
