Why is $e^x$ the only function that is its own derivative? I've heard that $f(x) = Ae^x$ is only function (both elementary and non-elementary) that satisfies the property $f(x)=\frac{df(x)}{dx}$. Is this true (and if it's true, is there a definitive way to prove it)?
 A: Any constant times $e^x$ also has this property.  
To see that these are the only examples, suppose we have a function $g(x)$ with $g'(x)=g(x)$.  Let $h(x)=g(x)e^{-x}$.  Note that $e^x$ is never $0$ so $h(x)$ is well-defined.  We compute $$h'(x)=g'(x)e^{-x}-g(x)e^{-x}=0\implies h(x)=\text {constant}$$
and we are done.
A: Let $y=f(x)$ thus the condition 
$$f’(x)=f(x)$$
is a differential equation and for the Theorem of Existence and Uniqueness an unique solution exists up to a constant.
Notably by separation of variables 
$$\frac{dy}{dx}=y\implies \frac{dy}{y}=dx\implies \log y = x+c$$
Which shows that the unique solution is the inverse of $\log$ function.
A: Let's do a "function research" for functions of the type $f(x) = f^{\prime}(x)$.
From $f(x) = f^{\prime}(x)$ it follows that $f$ should be infinitely often differentiable.
So, you may try to check whether there is a meaningful Taylor series around $x= 0$ describing such an $f$. So set $A = f(0)$ and let's try $A \neq 0$. So, you get
$$\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n = A \sum_{n=0}^{\infty}\frac{1}{n!}x^n  =  A \cdot E(x)$$.
Now you study $E(x)$ and find all the nice properties like $E(x+y) = E(x)E(y)$ etc. and realize that this function is uniquely determined and is usally written as $e^x$.
