From $e^n$ to $e^x$ Solve for $f: \mathbb{R}\to\mathbb{R}\ \ \ $        s.t. 
$$f(n)=e^n \ \ \forall n\in\mathbb{N}$$
$$f^{(y)}(x)>0 \ \forall y\in\mathbb{N^*} \ \forall x\in\mathbb R$$
Could you please prove that there exists an unique solution: $f(x)=e^x$?
(Anyway, this problem is not about fractional calculus)
$\mathbb N^*=\{1,2,3...\}, \ \mathbb N=\{0,1,2....\}$

How about try to construct a few functional spaces that intersect at one point?
Try Sard Theorem and Pre image Theorem.
 A: We can show that if
$$
f^{(n)} \ge 0\qquad{\rm(1)}
$$
everywhere, for each integer $n >0$, and $f(x)=e^x$ on at least four points of $\mathbb R$, then $f(x)=e^x$ everywhere. This will make use of Bernstein's theorem. I feel that there must be a more elementary proof though.
I'll start with the simpler case where (1) holds for all $n\ge0$, in which case it is only necessary to suppose that $f(x)=e^x$ at three points. As $\left(-\frac{d}{dx}\right)^nf(-x)=f^{(n)}(-x)\ge0$, by definition $f(-x)$ is completely monotonic. Bernstein's theorem means that we can write
$$
f(-x)=\int e^{-xy}\,\mu(dy)\qquad {\rm(2)}
$$
for some finite measure $\mu$ on $[0,\infty)$ and for all $x > 0$. By inversion of Laplace transforms, $\mu$ is uniquely defined. In fact, the statement (2) applies for all $x\in\mathbb R$ and we can argue as follows -- Applying the same argument to $f(K-x)$, for any fixed real $K$ extends (2) to all $x-K > 0$ and, letting $K$ go to $-\infty$, to all $x\in\mathbb R$. Changing the sign of $x$, for convenience,
$$
f(x)=\int e^{xy}\,\mu(dy).
$$
See also, this answer on MathOverflow (and the comments). Suppose that $\mu$ has nonzero weight outside of the set $\{1\}$. Multiplying by $e^{-x}$ and taking second derivatives
$$
\left(\frac{d}{dx}\right)^2e^{-x}f(x)=\left(\frac{d}{dx}\right)^2\int e^{x(y-1)}\mu(dy)=\int(y-1)^2e^{x(y-1)}\mu(dy) > 0.
$$
This means that $e^{-x}f(x)$ is strictly convex, contradicting the fact that it is equal to 1 at more than two points of $\mathbb R$. So, $\mu$ has zero weight outside of $\{1\}$ and $e^{-x}f(x)=\mu(\{1\})$ is constant.
I'll now return to the case where (1) holds for $n > 0$ and $f(x)=e^x$ on a set $S\subseteq\mathbb R$ of size four. By the mean value theorem, $f^\prime(x)=e^x$ holds for at least one point between any two points of $S$ and, hence, holds for at least three points of $\mathbb R$. So, using the proof above, $f^\prime(x)=e^x$ everywhere. Integrating, $f(x)=e^x+c$ for a constant $c$. Then, in order that $f(x)=e^x$ anywhere, $c$ must be zero.
A: Perhaps someone can formalize this for me. Because f(n) = e^n some derivative of f(x) must not equal e^x for them to be different functions. If the yth derivative of f(x) > e^x then f(x) > e^x for x > some number unless the y+zth derivative of f(x) < e^x in witch case f(x) < e^x for x > some number. For each derivative where that derivative of f(x) != e^x a higher derivative of f(x) that is > or < e^x must exist to compensate. Each compensation can only make f(x) < or > e^x for x > some number. Thusly f(n) cannot equal e^n for all n unless all derivatives of f(x) equal e^x meaning f(x) = e^x.
