E: The Significance of the Natural Logarithm Base in Mathematics

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I have problems understanding why e is such an important number in the world of mathematics. The number pi ,for instance, equals the area of the unit circle and is the ratio circumference/diameter. Why exactly is e that important?

I know that e is the base of the natural logarithm, but is this just by chance and definition? Could it just have been any other number?
 
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e = \lim_{n\rightarrow \infty} (1+\frac{1}{n})^{n}

Basically it is the limit of that function.
 
I still can't see why that fact makes the number one of the most important ones in maths. It's possible to set up many other terms whose limit hardly gets any attention at all, hey?
 
e is so important because the exponential function "is the most important function in mathematics" (Rudin, "Real and Complex Analysis"). Why is it so important?
  • It is the only function, within a multiplicative constant, that is its own derivative.
  • \pi can be defined in terms of e: "there exists a positive number number \pi such that e^{i\frac{\pi}2} = i and such that e^z = 1 \iff \frac z{2\pi i} is an integer" (Rudin again)
  • The trigonometric and hyperbolic functions can be written in terms of the exponential function.
 
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Just to add the fact that the exponential function is its own derivative within a multiplicative constant means that it turns up again and again as the solution to many differential equations.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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