Is a function with nowhere vanishing derivatives analytic? My question is the following: Let $f\in C^\infty(a,b)$, such that $f^{(n)}(x)\ne 0$, for every $n\in\mathbb N$, and every $x\in (a,b)$. Does that imply that $f$ is real analytic?
EDIT. According to a theorem of Serge N. Bernstein (Sur les fonctions absolument monotones, Acta Mathematica, 52 (1928) pp. 1–66) if $f\in C^\infty(a,b)$ and $f^{(n)}(x)\ge 0$, for all $x\in(a,b)$, then $f$ extends analytically in a ball centered at $(a,0)\in\mathbb C$ and radius $b-a$! 
 A: If $f$ is $C^{\infty}$ every derivative is continuous, so the hypothesis on $f$ implies that each derivative $f^{(n)}$ has constant sign.  Such functions were studied by S. Bernstein and called regularly monotonic.  In particular he proved in 1926 that a regularly monotonic function is real analytic.
This 1971 AMM article by R.P. Boas provides a proof, more history, and further results along these lines.  See also this 1975 PAMS article of J. McHugh.
A: Yes, any such function is analytic.
Assume contrary, let $f$ be such a function.
Note that if it is analytic at two points then it has to be analytic everywhere between. So by taking restriction, we may assume that the function is not analytic in any subinterval.
We can assume that $0$ is a point in the interval.
Assume the Taylor series of $f$ at $0$ converges in the $\varepsilon$-neighborhood of $0$.
Denote by $\bar f$ its sum.
The monotonicity of $f^{(n)}$ gives a bound on the error $f(x)-\bar f(x)$ on one side from $0$; it follows that $\bar f(x)$ converges to $f(x)$ if $0<x<\varepsilon$ or $\varepsilon<x<0$, a contradiction.
It remains to consider the case when the Taylor series of $f$ at $0$ diverges in any neighborhood of $0$. 
In this case, for any $\varepsilon >0$,
there is arbitrary large  $n$ such that $|f^{(n)}(0)|>\tfrac{n!}{\varepsilon^n}$.
Applying monotonicity of $f^{(n)}$ and integrating, we get that $|f^{(k)}(x)|>2^n$ for any $k\le n$ and some $-4{\cdot}\varepsilon<x<4{\cdot}\varepsilon$, a contradiction.
