Questions and Answers in MSE and two other references:
Updates.
Ad 2). The Fourier transform is a special case of the double-sided Laplace transform: $$ F(p) = \int_{-\infty}^{+\infty} e^{-pt}\,f(t) dt \quad \Longrightarrow \quad F(i\omega) = \int_{-\infty}^{+\infty} e^{-i\omega t} f(t)\,dt $$ The Fourier transform, in turn, is a generalization of the complex Fourier series: start with equation (20) in the Wolfram reference and read until the end.
Ad 5). In this reference
it is argued on page 23 (with obviously a typo in it) that the Boltzmann probability distribution
$f(E)$ must have the following form: $$ f(E_1) \times f(E_2) = h(E_1+E_2) $$ Let's elaborate on this a little bit: $$ f(E_1) \times f(E_2) = h(E_1+E_2) \quad \Longrightarrow \quad h(E) = h(E+0) = f(0)f(E) $$ Derivative: $$ h'(E) = f(0)f'(E) = \lim_{\delta\to 0} \frac{h(E+\delta)-h(E)}{\delta} = \lim_{\delta\to 0} \frac{f(E)f(\delta)-f(0)f(E)}{\delta} =\\ f(E)\,\lim_{\delta\to 0} \frac{f(\delta)-f(0)}{\delta} = f(E) f'(0) \quad \Longrightarrow \quad f'(E) = \frac{f'(0)}{f(0)} f(E) $$ Continuing in terms of the article: $$ \frac{df(E)}{f(E)} = \frac{-dE}{E_c} \quad \Longrightarrow \quad f(E) = A e^{-E/E_c} $$ Which is the Ansatz for the Boltzmann distribution.
Ad 3). According to Wikipedia, De Moivre's formula is: $$ \left[cos(x)+i\sin(x)\right]^n = \cos(nx) + i\sin(nx) $$ And this can be proved for any integer $n$ , quite independent of Euler's formula. It's rather the other way around: because of the pattern $\;f(x)^n = f(nx)\;$ , de Moivre's formula can be considered as a heuristics for Euler's formula.
But uniquesolution is quite right: Probably the most fundamental fact about it is that it is the only measurable function for which $f(x+y)=f(x)f(y)$ for all $x,y$
Can we mimic this behavior of $\,e^x$ with the function $f(x) = \cos(x)+i\sin(x)$ ? From trigonometry we know that: $$ \cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)\\ \sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y) $$ Hence: $$ f(x+y) = \cos(x+y) + i \sin(x+y) =\\ \left[\cos(x)\cos(y) - \sin(x)\sin(y)\right] + i \left[\sin(x)\cos(y) + \cos(x)\sin(y)\right] =\\ \left[\cos(x) + i \sin(x)\right]\left[\cos(y) + i \sin(y)\right] = f(x)f(y) $$ So our $f(x)$ behaves like an exponential function.