Ok, so our starting point is:
$\exp: \mathbb{R} \rightarrow \mathbb{R}$ is defined as the solution to
$$\begin{cases} f'=f \\
f(0)=1.
\end{cases}$$
Let's, first of all, conclude that $\exp(x)$ is strictly increasing starting from $0$.
- $\exp(x)$ is strictly increasing for $x>0$ .
It is clear that $\exp$ is $C^1$. This, together with the fact that $\exp'(0)=1$, allows us to take a neighbourhood of $0$ in which $\exp$ is strictly increasing.
Now, let $A:=\{x \in \mathbb{R}_{\geq 0} \mid \exp \text{ is strictly increasing in } (0,x) \}$. If $A$ were bounded, we could repeat the same argument as above near $\sup A$ to conclude that there exists $y> \sup A$ in $A$. Hence, $A$ is unbounded, and $\exp$ is strictly increasing on $(a,b)$.
Sidenote: This doesn't satisfy me, since a Calculus student may not be familiar with the concept of supremum.
- $\exp(x) \neq 0 \quad \forall x$ .
Suppose $\exp(x)=0$ for some $x \in \mathbb{R}$, and let $a$ be the $\sup$ of all such $x$ (which exists by the previous item).
Take $a+1$. By the mean value theorem, there exists $\xi$ between $a$ and $a+1$ such that $\exp(\xi)=\exp'(\xi)=\exp(a+1)$.
Applying the mean value theorem once again, we find a number $\zeta$ greater than $a$ such that $\exp(\zeta)=0$. Since $a$ is the $\sup$ of all numbers $x$ such that $\exp(x)=0$, we have a contradiction.
Sidenote: Again, supremum.
- $\exp(x+y)=\exp(x)\exp(y)$.
Let $y \in \mathbb{R}$, and $g(x)=\frac{\exp(x+y)}{\exp(x)\exp(y)}$. By the previous item, $g$ is well-defined. By the quotient rule, $g$ is constant. Since $g(0)=1$, it follows that $\exp(x+y)=\exp(x)\exp(y)$ for all $x$. Since $y$ is arbitrary, we have our result.
- $\exp$ is strictly increasing in the whole line.
From the last item, $\exp(-x)=\frac{1}{\exp(x)}$. It is clear that if $x>0$ and $y<0$, then $\exp(x)>\exp(y)$ (since $\exp(x)>1$ if $x>0$, by the first item). Therefore, it suffices to consider $x<y<0$. If $x<y<0$, then $-x>-y>0$, from which we conclude that $\exp(-x)>\exp(-y)$. By taking the inverse, we have that $\exp(x)< \exp(y)$.
- $\exp(x) \rightarrow_{x \to \infty} \infty$.
Note that the derivative of $\exp$ on $(0,\infty)$ is greater than $1$ for all points, due to the first item. Hence, $\exp(x)>x$, and the result follows from the Squeeze Theorem.
- $\exp: \mathbb{R} \rightarrow (0,\infty)$ is a bijection.
It is clear from the last two items, since $\lim\limits_{x \rightarrow -\infty} \exp(x)=\lim\limits_{x \rightarrow \infty} \exp(-x)=\lim\limits_{x \rightarrow \infty} \frac{1}{\exp(x)}=0$.
- $\lim\limits_{x \to \infty} \frac{x^n}{\exp(x)}=0$ for all $n \in \mathbb{N}$.
Induction proves that $\exp(x) \geq \frac{x^i}{i!}$ if $x \geq 0$ for all $i$. Hence, take any $n$. We then have that $\exp(x) \geq \frac{x^{n+1}}{(n+1)!}$, and then
$$\frac{x^n}{\exp(x)} \leq \frac{(n+1)! x^n}{x^{n+1}} = \frac{(n+1)!}{x},$$
and the result follows from the Squeeze Theorem.
- $\exp$, as defined by the differential equation, is unique.
Suppose there is another functiong $g$ which satisfies the differential equation. Consider $h=\frac{g}{\exp}$. It follows by the quotient rule that $h$ is constant. It is clear that $h(0)=1$. Hence, $g=\exp$.
- $\ln'(x)=1/x$
It follows from the expression of the derivative of the inverse function. We have that $\exp(\ln(x))=x$, hence $\exp'(\ln(x))\ln'(x)=1$, from which we get that $\exp(\ln(x))\ln'(x)=1$. Therefore, $\ln'(x)=1/x$.
- $\lim\limits_{x \to \infty} (1+x)^{1/x}=\exp(1).$
Taking $\ln$, we have that $\lim\limits_{x \to \infty} \ln\big((1+x)^{1/x}\big)=\lim\limits_{x \to \infty} \frac{1}{x} \ln(1+x)=\lim\limits_{x \to \infty} \frac{\ln(1+x)-\ln(1)}{x}=\ln'(1)=1.$ By continuity of $\ln$, we have that $\lim\limits_{x \to \infty} (1+x)^{1/x}=\exp(1).$
Drawbacks: There are two main problems which I see to this approach:
- The existence of a $\sup$ argument, which may not be suitable for Calculus students (as mentioned before).
- We don't prove existence of the solution, but I don't think that is a big issue.
