Is $\int f=f-1\iff f(\cdot)=e^{\cdot}$ proved this way correct? I saw this on math overflow and made me wonder, why does it work, is it rigorous, can we really factor like this, and where can we use similar tricks;
Let $\int$ denote $\int_0^x$
Then solve $$\int f=f-1 \iff 1=\left (1-\int\right )f\iff f=\left (1-\int\right )^{-1} 1$$
Then with geometric series 
$$f=\left(1+\int+\iint+\iiint+\cdots\right)1=1+\int_0^x1~dx+\int_0^x\int_0^x1~dx+\cdots=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots=e^x$$
 A: (Write $(T f)(x)=\int_0^x f(t) \, dt$, so that we don't get conflict with what $\int^k$ means)
I shall work on the space of continuous functions on $[0,R]$ completed using the supremum norm $\| f \| = \| f \|_{\infty} := \sup_{x \in [0,R]} |f(x)|$; call this $X$. It should be clear that $T$ maps $X \to X$.
Let $x>0$ be sufficiently small ($<1$ will do). Then $ \| T f \| \leqslant x \| f \| $, using the trivial bound $ \left| \int_a^b f \right| \leq (b-a) \sup_{a<x<b} |f(x)| $, and so $\| T^k f \| \leq x^k\| f \|$. (In fact, using the Cauchy formula for repeated integration, we can do better: $\| T^k f \| \leq (x^k/k!)\| f \|$, and then $x$ can be any real value. It should be easy for you to improve my calculation to work on the whole line with this.)
Now consider the operator sum
$$ S_n = I + \sum_{k=1}^n T^k, $$
where $I$ is the identity operator (your $1$); this is called the Neumann Series of $T$. We can prove this converges to the inverse of $(I-T)$: we have
$$ S_n (I-T)f = (I-T)S_n f = f - T^{n+1} f, $$
so
$$ \| S_n (I-T)f-f \| = \| T^{n+1} f \| < x^{n+1} f \to 0 $$
as $n \to \infty$. (Similarly for $(I-T)S_n f$.) Since $X$ is complete, we therefore conclude that $ S_{n} f \to (I-T)^{-1}f $.
The above shows that your calculation with the geometric series works, since $(T^k 1)(x) = x^k/k!$, and an argument with the contraction mapping theorem will show that $e^x$ is the unique such function.
