How do we know that C14 decay is exponential and not linear?
It's also worth noting that there is nothing special about atoms.
If you have any system where in every period of time an event has a certain chance of happening which only depends on internal effects of the object and no memory or communications with others - you will get the same decay curve. It's purely a matter of the statistics.
If you have a handful of coins and every minute toss them all and remove all the heads into a separate pile - the number of coins remaining in the hand will decay with a half-life of 1 minute.
What is special about carbon14 - and why it is useful for archeaology is that new carbon14 is being made all the time in the atmosphere, and while you are alive you take in this new carbon so the decays don't have any effect until you die. It's like tossing the coins, but while you are alive adding new random coins after each toss - but then when you die have somebody else start to remove the heads. If you assume you died with an equal number of heads and tails, you can work out how many tosses have happened since you died - and so how long ago the sample died.
We can show this by thinking about what is happening.
Suppose we have a set of $N$ nuclei that are all radioactive. Each of these nuclei has a chance of decaying, $\lambda$. In people lifetimes, some people live longer and some live shorter than others, but there is an average lifetime; this is what $\lambda^{-1}$ represents for nuclei.
Now how many nuclei, $\Delta N$, decay over some interval of time, $\Delta t$, can be represented mathematically by (remember $\Delta N < 0$ because nuclei are decaying), $$\Delta N = - \lambda N \Delta t$$
We can rewrite this in differential form as, $$\frac{d N}{N} = -\lambda dt$$
Using standard calculus, $$\int \frac{d N}{N} =- \int \lambda dt$$
We will find that, $$\ln N =- \lambda t + C$$
Standard algebraic manipulation gives,
$$ N = e^{-\lambda t}e^{C} $$
At $t=0$ we say that $N=e^C=N_0$, thus we usually write the equation: $$N = N_0 e^{-\lambda t}$$
Re: How do we know that C14 decay's exponentially compared to linear?
Experiment.
Re: Have there been any studies to verify this?
Yes. You can read a nice writeup of some of the early studies of C14 at the nobel prize website: http://static.nobelprize.org/nobel_prizes/chemistry/laureates/1960/libby-lecture.pdf
But to expand a bit on the underlying model as to why all radioactive decay (and spontaneous emission in general) is exponential in nature:
Radioactive decay is exponential is because the nuclei have no memory. Nuclei don't grow old like people - an 80-year-old man is very different than a 2-year-old boy, and their expected remaining lifetimes are very different. But all nuclei (until the moment they decay) are perfectly identical: they have no memory of when they were born. You can't distinguish a C14 atom that was created 10 seconds ago from one lucky atom that has managed to survive for 10000 years.
Consequently, every C14 nucleus has the same probability of decaying at a given moment in time. Thus, if $N$ is the number of remaining C14 atoms in your sample, $$\frac{\partial N}{\partial t} = -\Gamma \cdot N$$ where $\Gamma$ is some constant that describes how fast the atoms tend to decay. This equation, of course, gives rise to $N(t) = N_0 \cdot e^{-\Gamma t}$, which is just exponential decay.