Are there infinitely many integer-valued polynomials dominated by $1.9^n$ on all of $\mathbb{N}$? The original post is below. Question 1 was solved in the negative by David Speyer, and the title has now been changed to reflect Question 2, which turned out to be the more difficult one. A bounty of 100 is offered for a complete solution.
Original post. It follows from the prime number theorem and the periodicity properties $f(n+p) \equiv f(n) \mod{p}$ that for each $A < e$ there are only finitely many integer polynomials $f \in \mathbb{Z}[x]$ such that $|f(n)| < A^n$ for all $n \in \mathbb{N}$. On the other hand, for each $k \in \mathbb{N}$ the binomial coefficient $\binom{n}{k}$ is an integer-valued polynomial in $n$ bounded by $2^n$.
Question 1. Are there infinitely many integer polynomials with $|f(n)| < e^n$ for all $n \in \mathbb{N}$?
Question 2. Given $A < 2$, are there only finitely many integer-valued polynomials $f \in \mathbb{Q}[x]$ with $|f(n)| < A^n$ for all $n \in \mathbb{N}$?
 A: $\def\ZZ\mathbb{Z}$Question 1: No. Let $C>1$. I will show that there are only finitely many $f(x)$ in $\ZZ[x]$ so that $|f(n)| \leq C^n$ for all $n \in \ZZ_{\geq 0}$.
Choose $d$ large enough that, for all $k>d$, we have $k! > 2 C^k$. I claim that a polynomial with $|f(n)|<C^n$ is determined by its values on $0$, $1$, ..., $d$. Suppose, to the contrary, that $f(n) \neq g(n)$ but that they agree for $0 \leq n \leq d$. Let $k$ be the first integer where $f$ and $g$ disagree.
So $f(x)-g(x)$ is divisible by $x(x-1)(x-2) \cdots (x-k+1)$, so $f(x) - g(x) = x(x-1) \cdots (x-k+1) h(x)$ for some $h$ with integer coefficients. So $f(k) - g(k) = k! h(k) \equiv 0 \bmod k!$. 
But, by assumption, $|f(k)|$ and $|g(k)| < C^k < k!/2$. So it is impossible that $f(k) \neq g(k)$ and $f(k) \equiv g(k) \bmod k!$. This contradiction concludes the proof.
Question 2 is still stumping me. 
A: Question 2:
The constant $A$ can be brought down to $\sqrt 3$, and probably
a bit below that but not all the way down to $1+\epsilon$.
Instead of the polynomial $f(n) = {n \choose m}$, use a
finite difference of such polynomials, 
$$
f(n) = \sum_{i=0}^m (-1)^i {m \choose i} {n \choose m+i}.
$$
This is the $x^n$ coefficient of 
$$
\frac1{1-x} \left( \frac{x(1-2x)}{(1-x)^2} \right)^m
$$
and can be estimated by contour integration on $|x| = 3^{-1/2}$;
the maximum of $|f(n)|^{1/n}$ occurs near $n=3m$, for which
the critical points are at $x = (3 \pm \sqrt{-3}) / 6$.
Note that for $f(n) = {n \choose m}$ the generating function was
$\frac1{1-x} (x/(1-x))^m$, and the maximum occurred near $n=2m$, 
for which the critical point was at $x = 1/2$.
The factor $1-2x$ kills that maximum, and it seems that
using $x/(1-x)$ and $(1-2x)/(1-x)$ to the same power is optimal.
To reduce $A$ further, try to include also some power of $(3x^2-3x+1)/(1-x)^2$
to kill the new critical point.
