Hello! I would like to present a proof (sketch) of the following observation.

This equation is interpreted as a unit vector with an angle of \( x \) on the complex plane as in the picture. This should be clear upon observing that \( cos \) and \( sin \) form the real and imaginary parts respectively.

The essential property of \( e^x \) is that it is its :own derivative at any input, i.e \( \frac{d e^x}{ dx} = e^x \)

Similarly the essential property of \( i \) or any complex number for that matter is that upon multiplying two complex numbers :their measured angles add up for the final result. We should *want* a proof that explicitly utilizes this understood behavior.

The key idea is to decompose \(e^{ix}\) into the product of many understood terms- We can then imagine the angle and magnitude of \(e^{ix}\) ** directly**!

We first observe the function \( \bigl(1 + \frac{x}{n}\bigr)^n \). Its derivative is \( \bigl(1 + \frac{x}{n}\bigr)^{n - 1}\). (Due to chain rule and power rule cancelling each other out, you may visualize these :here). The derivative is off by exactly one term, and if we observe the same in the limit of n at infinity we would want to say that the function is equal to its derivative. In other words it satisfies the same essential property as \( e^x \) above.

\[ \lim_{n \to \infty} \bigl(1 + \frac{x}{n}\bigr)^n = e^x \]

This decomposition into infinite terms can be made rigorous (i.e ~played by the rules) with some work on binomial expansions. Also, the function itself :did not fall out of the sky-.

Now due to the convenient nature of complex valued calculus we may peer at

\[ e^{ix} = \lim_{n \to \infty} \bigl(1 + \frac{ix}{n}\bigr)^n \]

The right hand side is a product of complex terms, so we can try to calculate the final angle of the product and the final length of the product. This will let us see how exactly the magnitude and angle shift arises in \( e^{ix} \).

The magnitude of each of those \( \bigl( 1 + \frac{ix}{n} \bigr) \) terms can be calculated as \( \sqrt{1^2 + \frac{x}{n}^2}\) by pythagoras. Now- very neatly- in the product of n such terms the magnitude will be.

\[ \Biggl( \sqrt{1^2 + \left( \frac{x}{n} \right)^2} \Biggr)^n = \Biggl( 1+ \frac{x^2}{n^2} \Biggr )^{ \frac{n}{2}} \]

Therefore we can point out,

that for the magnitude of \( e^{ix} \) it is equal to

\[ \lim_{n \to \infty} \Biggl( 1+ \frac{\frac{x^2}{2n}}{\frac{n}{2}} \Biggr)^{ \frac{n}{2}} = \lim_{n \to \infty}e^{\frac{x^2}{2n} } = 1\]

In other words* *the \( e^{ix} \) *is the product of infinite terms such that their magnitudes multiply to 1.*

Now the angle of the same \( \bigl( 1 + \frac{ix}{n} \bigr) \) terms can be each calculated as \( \tan^{-1}(\frac{x}{n}) \). In the product of n such terms the angle of the result would be the sum of n such term's angles as:

\[ \tan^{-1}(\frac{x}{n}) + \tan^{-1}(\frac{x}{n}) + \tan^{-1}(\frac{x}{n}) + \cdots \]

The limit of this as we consider a product of infinitely more terms will be x. Intuitively we know that \( \tan^{-1}(t) \) is roughly \( t \) for smaller \(t\). And with \(\frac{x}{n}\) shrinking arbitrarily close to zero we find the final angle sum exactly approximating \[ \frac{x}{n} + \frac{x}{n} + \frac{x}{n} + \cdots = n * \frac{x}{n} = x \]

A nittier and grittier argument is available in :here.

But in essence we have that the final angle is equivalent to the \( x \) in \(e^{ix}\) in radians due to \(\tan^{-1}\).

The :Taylor series series expansion of \(tan^{-1}(x)\) is

\[ tan^{-1}(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} ... \]

\[ lim_{n \to \infty} \biggl( n * tan^{-1}(x/n) \biggr)= x - \frac{x^3}{3n^2} + \frac{x^5}{5n^4} - \frac{x^7}{7n^6} ... \]

Therefore as \( n \to \infty \) we find all terms disappear and the limit is exactly x.

In fact- the above function was first described in trying to understand compound interest-- which can be described as repeated multiplication of your amount in the bank with a growth factor. Trying to :maximize the compounding nature of the interest results in discovering \( e \).