From $e^n$ to $e^x$

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.