Is $e$ a coincidence?

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 $a$.

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(requires a bit more rigor to be done right)

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.

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 $\frac{d}{dx}e^x = e^x$.