Finding the function that is its own derivative

I saw this "derivation" some years ago but now I can't find it.  If anyone knows the source please let me know.

Let's say we want to find a function $f(x)$ for which
$$\frac{d}{dx} f(x) = f^\prime (x) = f(x)$$
Well, how about $f(x) = x$?
$$f(x) = x \implies f^\prime (x) = 1$$
No, because $f^\prime (x) \ne x$. OK, how about $f(x) = 1+x$?
$$f(x) = 1+x \implies f^\prime (x) = 1$$
That get's me a little closer.  I get the "$1$" back, but I need something whose derivative will give me back the $x$:
$$f(x) = 1+x+\tfrac{1}{2}x^2 \implies f^\prime (x) = 1 + x$$
Now I need something whose derivative will me back the $\tfrac{1}{2}x^2$:
$$f(x) = 1+x+\tfrac{1}{2}x^2 + \tfrac{1}{2 \cdot 3}x^3 \implies f^\prime (x) = 1 + x + \tfrac{1}{2}x^2$$
OK, so if I use infinitely many terms then this function fits the bill
$$f(x) = 1+x+\tfrac{1}{2}x^2 + \tfrac{1}{2 \cdot 3}x^3 + \cdots \tfrac{1}{n!}x^n \cdots  \implies f^\prime (x) = 1 + x + \tfrac{1}{2}x^2 + \cdots \tfrac{1}{(n-1)!}x^{n-1} \cdots$$
Can I write $f(x)$ in some compact form?  Well, $f(0) = 1$ and $f(1)$ will give me some number, which I'll call $e$.
$$f(1) = 1+1+\tfrac{1}{2} + \tfrac{1}{2 \cdot 3} + \cdots = e$$
You only need a few terms before $e$ has converged to the first two decimal places (2.717).

For $x$ = 2 you need a few more terms to get the same precision, but the number (7.382) is roughly 2.717 x 2.717, an agreement that quickly gets better with a few more terms
$$f(2) = 1+2+\tfrac{1}{2}4 + \tfrac{1}{2 \cdot 3}8 + \cdots = e^2$$
So
$$f(x) = e^x = 1+x+\tfrac{1}{2}x^2 + \tfrac{1}{2 \cdot 3}x^3 + \cdots \tfrac{1}{n!}x^n \cdots$$
and
$$\frac{d}{dx} e^x = e^x$$


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