Tao, in his Analysis II (3e) defines exponential as:
Definition 4.5.1 (Exponential function). For every real number $x$, we define the exponential function $\exp(x)$ to be the real number
$$
\exp(x) := \sum_{n = 0}^\infty\frac{x^n}{n!}.
$$
Then he defines $e$ as:
Definition 4.5.3 (Euler’s number). The number $e$ is defined to be
$$
e := \exp(1) = \sum_{n = 0}^\infty\frac{1}{n!} = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \cdots.
$$
Then is mentioned the proposition:
Proposition 4.5.4. For every real number x, we have $\exp(x) = e^x$.
That is,
$$
\sum_{n = 0}^\infty\frac{x^n}{n!} = \left(\sum_{n = 0}^\infty\frac{1}{n!}\right)^x.
$$
(For $x > 0$ and $r\in\mathbb R$, $x^r$ is defined to be the limit of the Cauchy sequence $(x^{q_n})_n$ where $(q_n)$ is a rational sequence converging to $r$.)
The proof is left as Exercise 4.5.3, which gives the following hint:
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 (Proposition 6.7.3) to be useful.
(Proposition 6.7.3 states just the basic laws of exponentiation.)
However, I am not even able to get off with the natural number case. Can someone help?
I am aware of the fact that $\lim_{n\to\infty}(1 + x/n)^n = \exp(x)$ (which Tao doesn't mention in the main text), which solves the problem: Just note that $\exp(x) = \lim_{n\to\infty}(1 + x/n)^n = \bigl(\lim_{n\to\infty}(1 + 1/n)^n\bigr)^x = \exp(1)^x = e^x$ (the second equality following from continuity of exponentiation in the base, which has already been proven in Vol I). But this is most probably not Tao has intended it to be solved.
Towards solution: Thanks to Anne Buval for (calming me down XD and) pointing me towards induction:
One notes that
\begin{align*}
\left( \sum_{n = 0}^\infty\frac{x^n}{n!} \right)
\left( \sum_{n = 0}^\infty\frac{1}{n!} \right)
& = \sum_{n = 0}^\infty\sum_{k = 0}^n\frac{x^k}{k!}\frac{1}{(n – k)!}\\
& = \sum_{n = 0}^\infty\frac{(x + 1)^n}{n!}\\
\end{align*}
where the first equality follows since both of the series (actually, just one is required) are absolutely convergent.
Best Answer
The answer is pretty straightforward indeed:
It is easily shown that $\exp(x + y) = \exp(x)\exp(y)$. (Use the fact that the series are absolutely convergent so that their Cauchy product converges to the product of their sums; and use binomial expansion.) Also note that $\exp(0) = 1$. Now it is a simple consequence of these two facts that $\exp(q) = e^q$ for any rational $q$. Now, since $\exp$ and $x\mapsto e^x$ are continuous, and $\mathbb Q$ dense in $\mathbb R$, conclude that $\exp(x) = e^x$ for all real $x$.