Characterization of the sinus I am looking for a simple proof of this result:  

If  $f\in C^{\infty}(\mathbb{R})$ fulfills $f'(0)=1$ and $|f^{(n)}(x)|\leq 1$  for any $x\in\mathbb{R}$ and any $n\in\{0,1,2,\ldots\}$, then $f(x)=\sin(x)$.

A proof (through the Paley-Wiener theorem and complex analysis) can be found here: 
http://www.math.sciences.univ-nantes.fr/~nicoleau/sinus.pdf
 A: This approach should do the job nicely. $f\in C^{\infty}(\mathbb{R})$ ensures $f\in L^2(-\pi,\pi)$, hence we may assume that
$$f(x) = k_0 + s_1 \sin(x)+c_1 \cos(x)+\sum_{n\geq 2}\left(c_n \cos(nx)+ s_n\sin(nx)\right) \tag{1}$$
holds for any $x\in(-\pi,\pi)$. Since $|f^{(n)}(x)|\leq 1$ we have $\int_{-\pi}^{\pi}f^{(n)}(x)^2\,dx \leq 2\pi$.
On the other hand, by Parseval's theorem, for any $m\geq 1$ we have
$$ \int_{-\pi}^{\pi}f^{(m)}(x)^2\,dx = \pi(s_1^2+c_1^2)+\pi\sum_{n\geq 2}n^{2m}(c_n^2+s_n^2) \tag{2}$$
hence in order that $\int_{-\pi}^{\pi}f^{(m)}(x)^2\,dx \leq 2\pi$ for any $m\geq 1$ we need to have
$$ \forall n\geq 2,\qquad c_n=s_n=0.\tag{3} $$
This proves$^{(*)}$ the only functions fulfilling our constraints are functions of the form $f(x)=k_0+s_1 \sin(x)+c_1\cos(x)$. By imposing $f'(0)=1$ we get $s_1=1$ and by further imposing $\left|f(x)\right|\leq 1$ we get $k_0=c_1=0$ as wanted.
$(*)$ On the interval $(-\pi,\pi)$. On the other hand the constraints imply that $f$ is an entire function, hence once $f$ is defined on $(-\pi,\pi)$ it can be prolongated in a unique way to the whole real line.
Or you may apply the same argument over and over, first on $L^2(-\pi,\pi)$, then on $L^2(-2\pi,2\pi)$, then on $L^2(-4\pi,4\pi)$, $\ldots$, concluding that $f(x)=\sin(x)$ over any interval of the form $(-2^N\pi,2^N\pi)$, i.e. on the whole real line.
In the comments below Bob Pego makes an important observation: in order to grant that the Fourier series of $f^{(n)}(x)$ is the formal $n$-th derivative of the Fourier series of $f(x)$ some assumption on the decay of $s_n,c_n$ has to be made. Luckily, the Paley-Wiener theorem grants that they have an exponential decay. Unluckily, we are not really dodging that crucial step.
The idea is freely stolen from the Turàn power sums inequality.
Problem solving is really instructing. Before looking at a proof of this fact I was believing it was wrong, and I tried constructing (non-working) counter-examples like $f(x)=\frac{\text{Si}(\text{Si}(\pi x))}{\text{Si}(\pi)}$. After many failed attempts, I convinced myself the claim was correct. Then I saw the proof through Paley-Wiener, and believed it was the most efficient way for proving the statement. The last step back I took was to exhibit this simple proof through Fourier series. Thank you, David Hilbert, because reflexive spaces are a bless.
A: This proof gives a comprehensive proof in the periodic case. It is not true that the periodized restriction of an analytic function on any bounded interval (even if the values of the function are the same at each endpoint of the segment) is smooth... 
 And it is not true that the use of Paley-Wiener Theorem gives the best way to deal with this exercise as well!
There are many "elementary" proofs of these results even perturbative one...
