Prove that $\exp(x) = e^x$ for every real number $x$. (Hint: first prove the claim when $x$ is a natural number. Then prove it when $x$ is an integer. Then prove it when $x$ is a rational number. Then use the fact that real numbers are the limits of rational numbers to prove it for all real numbers. You may find the exponent laws to be useful.)
The author defines $e: = \exp(1) = \sum_{n=0}^\infty 1/n!$. I can show that $\exp(x) = e^x$ when $x$ is an integer.
When $x$ is a rational number, let $x = \frac{n}{m}$ for $n, m \in \mathbb{N}$ (if $x$ is negative, I can use the result that $\exp(-x) = 1/\exp(x)$). Then, $\exp(n/m) = \exp(1/m + ... + 1/m) = \exp(1/m) \cdot\cdot\cdot \exp(1/m)$. Thus, the problem is reduced to showing $\exp(1/m) = e^{1/m}$. How can I show this?
You could just take $n = m$. Then $$e = \exp(1) = \exp(m/m) = \exp(1/m)^m$$ and raise each side to the $1/m$ power.