Derivation of the Exponential Nature of $e^x$ Presumably, the transcendental number $e$ was first found by taking the power series solution to the (arguably most fundamental) differential equation $f'(x)=f(x)$, with the initial condition $f(0)=1$ and then plugging in $x=1$. My question is, is there any way, other than using the equivalent of the multinomial theorem for power series to demonstrate that $f(nx)=f^n(x)$ for all $n$ and $x$ (i.e., that $f(x)$ is an exponential function) other than this laborious and unaesthetic method? Intuitively, I would expect the solution to be an exponential function but is there a better, somewhat rigorous demonstration of this simple fact?

I believe that $e$ was first found as $$ e=\lim_{n\to\infty}\left(1+\frac1n\right)^n\tag{1} $$ perhaps arising from compound interest problems and the like.

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However, if we start with $f'(x)=f(x)$, then we have for any fixed $y$ $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\frac{f(x+y)}{f(x)} &=\frac{f'(x+y)f(x)-f(x+y)f'(x)}{f(x)^2}\\\ &=\frac{f(x+y)f(x)-f(x+y)f(x)}{f(x)^2}\\\\[6pt] &=0\tag{2} \end{align} $$ Therefore, $\frac{f(x+y)}{f(x)}$ is constant in $x$. In particular, $$ \frac{f(x+y)}{f(x)}=\frac{f(y)}{f(0)}\tag{3} $$ which, if we have $f(0)=1$, yields $$ f(x+y)=f(x)f(y)\tag{4} $$ Equation $(4)$ easily gives $f(nx)=f(x)^n$ by induction.