Is $e$ a coincidence? $e$ has many definitions and properties. The one I'm most used to is 
$$\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^n $$
If someone asked me (and I didn't know about $e$):

Is there a constant $c$ such that the equation $\frac{d}{dx}c^x=c^x $ is true for all $x$?

Then I'd likely answer that: 

I doubt it! That would be a crazy coincidence.

I'm curious, is it a coincidence that there is a constant that makes this true?
 A: One way to see that such a function has to exist is to notice that the derivative of an exponential function is proportional to the original function:
$$\frac{d}{dx} c^x = \lim_{h\to 0}\frac{c^{x+h}-c^x}{h} = c^x\cdot \lim_{h\to 0} \frac{c^h-1}{h}$$
The constant of proportionality is given by the result of the limit on the right. For $2^x$, you can numerically verify that $\frac{d}{dx}2^x \approx 0.69\cdot 2^x$. Similarly, $\frac{d}{dx} 3^x\approx 1.10\cdot 3^x$. In the former case, the constant is less than $1$, and in the latter case, it's greater than $1$. So it's not unreasonable to suppose that there is a number $e$ in between $2$ and $3$ that makes the constant exactly $1$, i.e. such that $\frac{d}{dx}e^x = e^x$. 
A: First, note that you only need to know that the derivative is $1$ at $x=0$. This is because, for general $x$:
$$\lim_{h\to 0} \frac{c^{x+h}-c^x}{h}=c^x\lim_{h\to 0}\frac{c^h-1}{h}$$
So if the derivative of $f(x)=c^x$ exists at $x=0$ then $f'(x)=c^xf'(0)$ for any $x$.
Now, if $f(x)=2^x$ has $f'(0)\neq 0$ then a linear change of variable has some $g(x)=2^{x/f'(0)}$ for some $a$ has $g'(0)=1$. Let $e=g(1)$, and $g(x)=e^x$.
So the only thing you really need to believe to assert such $e$ exists is that for some $a$, $a^x$ has a non-zero derivative at $x=0.$
A: It is no coincidence.  Note that
$$\frac d{dx}a^x=\lim_{h\to0}\frac{a^{x+h}-a^x}h=a^x\lim_{h\to0}\frac{a^h-1}h$$
So we're really trying to solve the following:
$$\lim_{h\to0}\frac{a^h-1}h=1$$
Assuming the limit exists, we consider the limit from the right hand side with $h\mapsto1/n$ to get
$$\lim_{n\to+\infty}n(a^{1/n}-1)=1$$
Consider a sequence of numbers $a_n$ that satisfy the following equality:
$$n((a_n)^{1/n}-1)=1\\(a_n)^{1/n}-1=\frac1n\\(a_n)^{1/n}=1+\frac1n\\a_n=\left(1+\frac1n\right)^n$$
It thus stands to reason that $a=\lim_{n\to\infty}a_n=e$ is the right choice of $a$.

(requires a bit more rigor to be done right)
A: It is not a coincidence.
$$e^x = \lim_{n \rightarrow \infty} \Big(1+ \frac{x}{n} \Big)^n$$
$$ \frac{d}{dx} \Big(1+ \frac{x}{n} \Big)^n = \Big(1+ \frac{x}{n} \Big)^{n-1}$$
Then Euler would say "it is obvious from above that"
$$ \frac{d}{dx} e^x = e^x$$
We understand that it is not obvious that
$$\lim_{n \rightarrow \infty} \frac{d}{dx} \Big(1+ \frac{x}{n} \Big)^n = \frac{d}{dx} \lim_{n \rightarrow \infty}  \Big(1+ \frac{x}{n} \Big)^n$$
I can prove it for you. But I hope I have answered your meta question.
A: You have several good answers (which I have upvoted). Here's a partial answer (similar to @florence 's but without the definition of the derivative) that addresses this question:

Is there a constant $c$ such that the equation $\frac{d}{dx}c^x=c^x$ is true for all
  $x$?

without getting to the definition of $e$.
If you sketch the graph of $y=2^x$ (or think about population doubling) you will note that its slope is roughly proportional to its value - but it's never as steep as it is tall. The proportionality constant is less than $1$. If you do the same for $y=3^x$ you'll see that it's always steeper than it is tall. That makes it at least plausible that there's a constant $c$ with value between $2$ and $3$ that satisfies the differential equation. The particular value of that constant may be "coincidence" but its existence isn't.
A: If we define, for $p>0$
$$
\log p=\int_1^p \frac{1}{x}\,dx
$$
and consider $p>1$ for simplicity, then we get an approximation of $\log p$ by dividing the interval $[1,p]$ in $n$ parts in geometric progression: letting $r=\sqrt[n]{p}$, we consider
$$
r^0=1 \quad r\quad r^2\quad \dots \quad r^n=p
$$
The “upper area” in the interval $[r^{k-1},r^k]$ is
$$
\frac{1}{r^{k-1}}(r^k-r^{k-1})=r-1
$$
so the upper sum is
$$
n(r-1)=n(\sqrt[n]{p}-1)
$$
Let's try and find the number $e$ for which $\log e=1$, by solving
$$
n(\sqrt[n]{e_n}-1)=1
$$
We get
$$
e_n=\left(1+\frac{1}{n}\right)^{\!n}
$$
which defines a sequence converging to the desired number $e$:
$$
e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{\!n}
$$
Why is this number important? Well, if $\exp$ is the inverse function of $\log$, we have, with an easy induction,
$$
\exp k=(\exp 1)^k
$$
and, by definition, $\exp 1=e$, so
$$
\exp k=e^k
$$
We can show similarly, that $\exp(a/b)=\sqrt[b]{e^a}=e^{a/b}$ for any integers $a$ and $b$ (with $b>0$). Thus it makes sense to write
$$
\exp x=e^x
$$
for every real $x$. The property $\exp(x+y)=(\exp x)(\exp y)$ follows directly from $\log(pq)=\log p+\log q$.
By differentiating $\log(\exp x)=x$, we deduce
$$
\frac{\exp' x}{\exp x}=1
$$
In other words $\exp' x=\exp x$.
A: Just to add how these definitions of $e$ are matched together. You can call this an overkill. I use Taylor polynomials around $x=0$ and then Taylor series to show that $e$ satisfies the equation.
We first note that $c=0$ will not work (why?). It is obvious that the function $f(x)=c^x$ is infinitely many times differentiable. So, $f(0)=f'(0)=...=f^{(n)}(0)=1$. So the Taylor polynomial around $x=0$ is given as:
\begin{align}
T_n(x)=1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!} 
\end{align}
With remainder term:
\begin{align}
R_n(x) = \frac{c^A}{(n+1)!}x^{n+1}
\end{align}
for $A$ between $x$ and $0$. It is obvious that $\vert R_n(x) \vert \to 0$ as $n \to \infty$ for all $x\in \mathbb{R}$. So the function can be expressed in a Taylor series and it is equal to the function on all of $x\in\mathbb{R}$:
\begin{align} 
f(x)=1+\sum_{k=1}^{\infty} \frac{x^k}{k!}  
\end{align}
So:
\begin{align}
f(1)=c=1+\sum_{k=1}^{\infty} \frac{1}{k!}  
\hspace{25pt} (1)
\end{align}
If your definition of $e$ was as $(1)$, then we would be finished. But if it was:
\begin{align}
e=\lim\limits_{n\to \infty}\left(1+\frac 1 n\right)^n \hspace{25pt} (2)
\end{align}
Then we should do more work. In case you wonder how to do that, define:
\begin{align}
e_n = \left( 1 +\frac 1 n\right)^n
\end{align}
Now we have:
\begin{align}
e_n &= \left( 1 +\frac 1 n\right)^n\\ 
&= \sum_{k=0}^n \binom{n}{k}\frac{1}{n^k} \\
&= 1+ \sum_{k=1}^n \binom{n}{k}\frac{1}{n^k} \\
&= 1+  \sum_{k=1}^n \frac{1}{k!}\frac{n \cdot (n-1) \cdot ... \cdot (n-k+1)}{n\cdot n ... \cdot n}\\
&\leq 1+  \sum_{k=1}^n \frac{1}{k!}\frac{n \cdot n \cdot ... \cdot n}{n\cdot n ... \cdot n}\\
& = 1+  \sum_{k=1}^n \frac{1}{k!} 
\end{align}
Taking $n \to \infty$ we get: $e\leq f(1)=c$. We can do something similar to get: $e\geq f(1)=c$. And that implies $c=e$ as desired. 
You see now how some of the definitions of $e$ imply the other.
Edit:
@Alex M. asked me to add an elaboration on the part  $e\geq f(1)=c$.
Note that for $n\geq m$ we have:
\begin{align}
e_n =& 1+\sum\limits_{k=1}^n \binom{n}{k} \frac{1}{n^k}\\
=& 1+  \sum_{k=1}^n \frac{1}{k!}\frac{n \cdot (n-1) \cdot ... \cdot (n-k+1)}{n\cdot n ... \cdot n}\\
\geq& 1+  \sum_{k=1}^m \frac{1}{k!}\frac{n \cdot (n-1) \cdot ... \cdot (n-k+1)}{n\cdot n ... \cdot n} \hspace{15pt}\text{ (since } n\geq m)\\ 
=& 1+\sum_{k=1}^m \frac{1}{k!} \cdot \frac{n}{n} \cdot \frac{n-1}{n} \cdot ... \cdot \frac{n-k+1}{n}
\end{align}
Now we take $n \to \infty$ to obtain:
\begin{align}
e\geq 1+\sum_{k=1}^m \frac{1}{k!}
\end{align} 
Now we take $m \to \infty$ to get:
\begin{align}
e\geq 1+\sum_{k=1}^\infty \frac{1}{k!} = f(1)=c
\end{align}
