Is it possible to interpolate $e^n$ in more than one way? The most basic definition of exponentiation is repeated multiplication, $$e^n = e \cdot e \cdot  \cdot  \cdot  \cdot  e$$ $n$ times 
However, if $n$ is a rational number such as $2.4$, this definition breaks down. One way we've been able to interpolate the function $e^x$ to include all real values is by using infinite series $$e^x = \sum_{n = 0}^{\infty} \dfrac{x^n}{n!}$$ 
My question is, is it possible to find another function $f$ which is smooth over all the reals and agrees on integer values with $e^x$ but disagrees on rational values?
If it exists, could we use such a function to completely rewrite/overhaul a huge portion of all the math that exists today?
 A: I believe there are tons of such functions. One of them is:
$$e^x \cos (2\pi x)$$
A: There are a couple of interesting ways to tackle this.  As @KayK. shows us, for any periodic function $f(x)$ over the integers not identically equal to one, then $f(x)e^x$ will satisfy your problem.
As @fleablood mentions, we can get a more interesting function.  Unlike the previous example, we can not only keep up with $e^x$ over integers, but we can also maintain some properties of $e^x$ and equal $e^x$ over the rationals.  For example, consider any function that satisfies
$$f(x+y)=f(x)f(y),~~f(1)=e,~~f(x)>0\forall x\in\mathbb R$$
It then follows trivially that $f(n)=e^n$ for $n\in\mathbb N$.  What's more, we can even show that $f(x)=e^x$ for all $x\in\mathbb Q$.  What we can't show is that $f(x)=e^x$ for $x\in\mathbb R$.  Indeed, assign $f(\pi)=3$ and you'll find this doesn't affect $f(x)$ over the rationals.
However, our previous attempt produce a discontinuous function.  (You'll find with the condition of continuity, $f(x)=e^x$ for all $x\in\mathbb R$.)
