Shorter proof for the existence of $e$ where $de^x/dx=e^x$? Without using a priori knowledge of $e$ or the natural logarithm, I'm looking for a shorter proof to the statement $$\exists e>0:\forall x\in\mathbb{R}:\frac{d}{dx}e^x=e^x.\tag{1}$$
The shortest I have come up with is this, in outline form:


*

*Derive the Maclaurin series $f(x)=\sum_{n=0}^\infty\frac{x^n}{n!}$ from the 2 conditions $f(x)=f'(x)$ and $f(0)=1$.

*Prove absolute convergence for all $x\in\mathbb{R}$ using a standard convergence theorem.

*Derive $f(x+y)=f(x)f(y)$ from multiplying out $f(x+y)=\sum_{n=0}^\infty\frac{(x+y)^n}{n!}$, using the binomial theorem, and re-arranging the infinite series into the product of $\sum_{n=0}^\infty\frac{x^n}{n!}$ and $\sum_{n=0}^\infty\frac{y^n}{n!}$.

*Show $\forall x\in\mathbb{R}:0<f(x)$ from $f(x)=f(x/2)^2\ge0$, and $f(x)f(-x)=1$ so $f(x)\neq0$.

*$\forall x\in\mathbb{N}:\forall y\in\mathbb{R}:f(xy)=f(y)^x$ from induction on $x$.

*$\forall x\in\mathbb{Z}:\forall y\in\mathbb{R}:f(xy)=f(y)^x$ from $f(-xy)=1/f(xy)=1/f(y)^x=f(y)^{-x}$ for $x\in\mathbb{Z}_+$.

*$\forall x\in\mathbb{Q}:\forall y\in\mathbb{R}:f(xy)=f(y)^x$ from $f(ym/n)=f(y/n)^m=f(y/n)^{mn/n}=f(yn/n)^{m/n}=f(y)^{m/n}$ for $(m,n)\in\mathbb{Z}\times\mathbb{Z}_+$.

*$\forall x\in\mathbb{R}:\forall y\in\mathbb{R}:f(xy)=f(y)^x$ from $f(xy)=\lim_{i\rightarrow\infty}f(x_iy)=\lim_{i\rightarrow\infty}f(y)^{x_i}=f(y)^x$ for any rational sequence $x_i$ that converges to $x$.

*Finally, define $e\equiv f(1)$. $f(x)=f(1)^x=e^x$.


Is there a shorter proof?
 A: I like to start at
$f(x+y)
= f(x)f(y)
$,
since that's why we use
the exponential function.
Also,
as Ellington didn't say,
it don't mean a thing
if it ain't got
that derivative,
so I assume that
$f$ is differentiable.
Putting $y=0$,
$f(x)f(0) = f(x)$,
so $f(0) = 1$.
Then
$\begin{array}\\
f(x+h)-f(x)
&=f(x)f(h)-f(x)\\
&=f(x)(f(h)-1)\\
\text{so}\\
\dfrac{f(x+h)-f(x)}{h}
&=f(x)\dfrac{f(h)-1}{h}\\
&=f(x)\dfrac{f(h)-f(0)}{h}\\
\text{Letting } h \to 0\\
f'(x)
&=f(x)f'(0)\\
\end{array}
$
From here on,
other properties follow
as above.
The regular exponential function
is characterized
by having
$f'(0) = 1$.
A: Here is seven(six) steps:


*

*Define $\log (x)$ by $\log(1) = 0$ and $ \log(x)' = \frac{1}{x}$.

*Prove $\log(xy) = \log(x) + \log(y)$ by considering the function $f(x) = \log(xy) - \log(x) - \log(y)$ which can be shown to have derivative zero and to be equal to zero at $x=1$.

*Define the exponential using a power series.

*Prove the exponential series converges.

*Use another differentiation argument to prove $\log(\exp(x)) = x$. Do this by defining $g(x) = \log(\exp(x)) - x$ then showing $g' = 0$.

*Define $x^y \triangleq \exp(y \log(x))$ as is standard.

*Notice $\exp (1)^x$ which is defined to be $\exp(x \log(\exp(1)))$ is equal to $\exp(x 1) = \exp(x)$.


Part of the reason this proof has to have so many steps is that in order to prove the number $e$ exists we must first define what the function $e: \mathbb R \rightarrow \mathbb R$  $ x \mapsto e^x$ should mean.
A: OP assumes that the function $a^{x}$ is defined for all $a, x \in \mathbb{R}$ and $a > 0$ such that usual properties of the exponents hold. And we are expected to show that there is a unique number $e$ such that $$\frac{d}{dx}e^{x} = e^{x}$$ for all $x \in \mathbb{R}$.
This is easy (because the biggest hurdle is to define $a^{x}$ without any use of $e$ and this has been assumed to be done somehow). Let $f(x) = a^{x}$ and then we have
\begin{align}
f'(x) &= \lim_{h \to 0}\frac{a^{x + h} - a^{x}}{h}\notag\\
&= \lim_{h \to 0}\frac{a^{x}a^{h} - a^{x}}{h} \text{ (via property of exponents)}\notag\\
&= a^{x}\lim_{h \to 0}\frac{a^{h} - 1}{h}\tag{1}
\end{align}
The challenge is to analyze the limit $$\lim_{h \to 0}\frac{a^{h} - 1}{h}$$ for all $a > 0$ and this we do now.
Using the inequality (which can be proven algebraically) $$\frac{a^{r} - 1}{r} > \frac{a^{s} - 1}{s}\tag{2}$$ for $r > s > 0, a > 1$ it is seen that $g(h) = (a^{h} - 1)/h$ is strictly increasing function of $h$ for $a > 1, h > 0$ and clearly $g(h) > 0$ for $h > 0$ so that $\lim_{h \to 0^{+}}g(h)$ exists. Further putting $h = -k$ we can see that $$\lim_{h \to 0^{-}}g(h) = \lim_{k \to 0^{+}}\frac{a^{-k} - 1}{-k} = \lim_{k \to 0^{+}}\frac{a^{k} - 1}{k}\cdot\frac{1}{a^{k}} = \lim_{h \to 0^{+}}g(h)$$ and therefore $$f'(0) = \lim_{h \to 0}g(h) = \lim_{h \to 0}\frac{a^{h} - 1}{h}$$ exists. Clearly this limit is dependent on the value of $a$ and let's denote it by $L(a)$.
We then have $L(1) = 0$ obviously and $L(a)$ exists and is non-negative for $a > 1$. If $0 < a < 1$ then we can see easily prove (by setting $b = 1/a$) that $L(a) = -L(1/a)$ so that $L(a) = \lim_{h \to 0}g(h)$ exists for all $a > 0$. Using further inequality (which can be proven algebraically) $$a^{h - 1}(a - 1) \leq \frac{a^{h} - 1}{h} \leq a - 1\tag{3}$$ for $h > 0, a > 1$ we can see that that $$L(a) = \lim_{h \to 0}\frac{a^{h} - 1}{h} \geq \frac{a - 1}{a}$$ for $a > 1$. Thus $L(a) > 0$ for $a > 1$ and since $L(a) = -L(1/a)$ it follows that $L(a) < 0$ for $0 < a < 1$.
Using the definition of $L(a)$ we can easily show that $$L(ab) = L(a) + L(b), L(a/b) = L(a) - L(b)\tag{4}$$ for positive $a, b$. From the second relation above we see that $L(a)$ is a strictly increasing function of $a$. Again from inequality $(3)$ we can see that $$\frac{a - 1}{a}\leq L(a) \leq a - 1$$ for $a > 1$ and hence by squeeze theorem $$\lim_{a \to 1^{+}}\frac{L(a)}{a - 1} = 1\tag{5}$$ If $a \to 1^{-}$ we can put $b = 1/a$ and use $L(a) = -L(b)$ to see that $$\lim_{a \to 1^{-}}\frac{L(a)}{a - 1} = 1$$ and thus we have finally $$\lim_{x \to 1}\frac{L(x)}{x - 1} = 1 = \lim_{x \to 0}\frac{L(1 + x)}{x}\tag{6}$$ From the above equation and $L(a/b) = L(a) - L(b)$ it can be easily seen that $L(x)$ is continuous and differentiable on $(0, \infty)$ and $L'(x) = 1/x$.
Further we know that $L(2) > 0$ and using $(4)$ we can get $$L(2^{n}) = nL(2), L(2^{-n}) = -nL(2)$$ so that range of $L(x)$ is $(-\infty, \infty)$. Since $L(x)$ is strictly increasing there is a unique number $e > 1$ such that $L(e) = 1$. Our job is now done. In the beginning we had shown that if $f(x) = a^{x}$ then $f'(x) = f(x)L(a)$ and hence if $f(x) = e^{x}$ then $f'(x) = f(x)L(e) = f(x)$. The function $L(x)$ is traditionally denoted by $\log x$.
The content of this answer is largely based on my blog post.
