I can't remember where I first saw this: does anybody recognise it?

Let $I$ be the operator, from $C^0(\mathbb{R})$ to itself, which takes $f(x)$ to $\int_0^xf(z)dz$.

Since the exponential function $e(x)$ is it's own derivative, we integrate both sides to get
$e(x) = I(e(x)) + 1$. Regarding $1$ as the identity operator, we can rearrange to get
$$(1-I)e(x) = 1,$$ and hence
$$e(x) = \frac{1}{1-I}1 = (1 + I + I^2 + \cdots)1 = 1 + x + \frac{x^2}{2} + \cdots.$$