How does one prove that $e$ exists? In my calculus class, $e$ was defined to be the number such that $\frac{d}{dx}e^x = e^x$.
From the definition of the derivative, we have
\begin{align*}
\frac{d}{dx}a^x &= \lim_{h \to 0} \frac{a^{x+h} - a^x}{h}\\
&= a^x \lim_{h \to 0} \frac{a^h - 1}{h}
\end{align*}
Thus $e$ is the number such that
$$
\lim_{h \to 0} \frac{e^h - 1}{h} = 1
$$
But how is it proven that there exists such number?
 A: Let $a>0$ and $a\ne1$. First we have to prove the existence of  $\displaystyle \lim_{h\to0}\frac{a^h-1}{h}$.
Assume that $r>1$ and let $f(x)=x^r-rx+r-1$ for $x>0$. Then
$$f'(x)=r(x^{r-1}-1)\begin{cases}<0 &\text{if }0<x<1\\
=0 &\text{if }x=1\\
>0 &\text{if }x>1 \end{cases}$$
Therefore, $f$ attains its absolute minimum at $x=1$. So  for all $x>0$, we have
$$f(x)\ge f(1)=0$$
$$x^r\ge rx+1-r$$
So when $r>1$ and $h>0$, $\displaystyle\frac{a^{rh}-1}{rh}\ge\frac{ra^h+1-r-1}{rh}$ and hence
\begin{align}
\frac{a^{rh}-1}{rh}-\frac{a^h-1}{h}\ge0
\end{align}
When $r>1$ and $h<0$, $\displaystyle\frac{a^{rh}-1}{rh}\le\frac{ra^h+1-r-1}{rh}$ and hence
\begin{align}
\frac{a^{rh}-1}{rh}-\frac{a^h-1}{h}\le0
\end{align}
Therefore, $\displaystyle \frac{a^h-1}{h}$ is an increasing function in $h$. As it is bounded below by $0$ on $(0,\infty)$, $\displaystyle \lim_{h\to0^+}\frac{a^h-1}{h}$ exists.
When $h<0$,
$$\frac{a^h-1}{h}=a^h\left(\frac{a^{-h}-1}{-h}\right)$$
As $\displaystyle \lim_{h\to0^-}a^h$ exists and equals $1$, $\displaystyle \lim_{h\to0^-}\frac{a^h-1}{h}$ exists and $\displaystyle \lim_{h\to0^-}\frac{a^h-1}{h}= \lim_{h\to0^+}\frac{a^h-1}{h}$.
Therefore, $\displaystyle \lim_{h\to0}\frac{a^h-1}{h}$ exists.
Now we are ready to prove that there exists an $e$ such that $\displaystyle \lim_{h\to0}\frac{e^h-1}{h}=1$.
Define $e=a^\frac{1}{k}$, where $\displaystyle k=\lim_{h\to0}\frac{a^h-1}{h}$. Then
\begin{align}
\lim_{h\to0}\frac{e^h-1}{h}&=\lim_{h\to0}\left(\frac{a^\frac{h}{k}-1}{\frac{h}{k}}\cdot \frac{1}{k}\right)\\
&=k\cdot\frac{1}{k}\\
&=1
\end{align}
This number $e$ is unique. Indeed, if $b>0$ and $\displaystyle \lim_{h\to0}\frac{b^h-1}{h}=1$, then we can prove that $b=e$.
Let $p=\log_eb$. Then $b=e^p$.
\begin{align}
\lim_{h\to 0}\frac{b^h-1}{h}-\lim_{h\to 0}\frac{e^h-1}{h}&=1-1\\
\lim_{h\to 0}\frac{e^{ph}-e^h}{h}&=0\\
\lim_{h\to 0}\left[(p-1)e^h\cdot\frac{e^{(p-1)h}-1}{(p-1)h}\right]&=0\\
(p-1)(1)(1)&=0\\
p&=1
\end{align}
Hence $b=e$.
A: It depends on how you define $e$. In some senses, you could define $$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} $$ and then from here set $x=1$ and you can prove that $e$ has all the properties we know and love. You can also show that this is equivalent to the limit definition of $e$. 
It exists because its definition is based on concepts that are already well-defined (and the power series is convergent for all $x \in \mathbb{R}$ and so defines a function on $\mathbb{R}$). 
A: Here's a proof for the alternative form $e := \lim_n (1+1/n)^n$.
Note first that the binomial theorem yields
$$e = \lim_n \sum_{k=0}^n \binom{n}{k} \left ( \frac{1}{n} \right )^k$$
and the elementary bounds $\frac{n^k}{k^k} \le \binom{n}{k} \le \frac{n^k}{k!}$ yield respective bounds
$$\lim_n \sum_{k=0}^n \frac{1}{k^k} \le e \le \lim_n \sum_{k=0}^n \frac{1}{k!}$$
Fixing any $n > 1$ and avoiding the limit for the lower bound gives the weaker bound $2 < e$, while noting that $k! \ge 2^{k-1}$ for $k \ge 1$ gives the weaker bound $e \le 1 + \lim_n \sum_{k=0}^n 2^{-k} = 3$, i.e., $2 < e \le 3$.
(Note that the sophomore's dream has made an appearance here.)
