Let us take the exponential function $\lambda^z$ where $0 < \lambda < 1$. There are many great uniqueness conditions this holomorphic function satisfies. For example, it is the only function holomorphic in the right half plane, bounded in the right half plane, that interpolates $\lambda^{z} \Big{|}_{\mathbb{N}}$. It is also the only function to satisfy the differential equation $\frac{d}{dz}f(z) = \log(\lambda)f(z)$ when $f(0) =1$. Also it is the only function such that $f(z)f(w) = f(z+w)$ and $f(1) = \lambda$ and $f:\mathbb{R}^+\to\mathbb{R}$. It is also the only function holomorphic such that $f(z)f(w) = f(z+w)$ and $f(1) = \lambda$ and $f$ has a period of $2 \pi i /\log(\lambda)$.

I'm asking this question because of this similar question on tetration An explicit series representation for the analytic tetration with complex height However when approaching the uniqueness of this tetration function on the real positive line, what is barring our path is a lemma (if you will) about the exponential function. This lemma is rather simple to state, and leads me to believe there must be a solution of this question.

I am not conjecturing this lemma to be true by calling it a lemma (I have a feeling its true); it'd just be nice to see a proof in the positive. If someone has references or suggestions or even a straight proof about why this must be true/not true, that'd be outstanding. The lemma has to do with completely monotonic functions, and is rather tricky. It can be stated plainly as

If a function $F$ is analytic on $\mathbb{R}^+$ and satisfies

$$F(1) =\lambda$$ $$\lambda F(x) = F(x+1)$$ $$(-1)^n\frac{d^n}{dx^n}F(x) >0$$

must $F(x) = \lambda^x$?

This can also be stated, if $F(x) = \lambda^{x}\phi(x)$ for some $1$-periodic function $\phi(x)$, and $(-1)^n\frac{d^n}{dx^n}F(x) > 0$, must $\phi(x)$ be constant?

It can also be stated in the manner I think is most likely to have an answer ready at hand. I also believe this to be the most valuable way of stating the question.

Letting $F$ be analytic on $\mathbb{R}^+$, if $F(1) = e$ and $$\frac{d^n}{dx^n}F(x) > 0$$ $$e\cdot F(x)= F(x+1)$$ must $F(x) = e^{x}$?