<s>[INST] 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?
 [/INST] 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}
</s>