n-th derivative where $n$ is a real number? We know that 
$$\frac{d^n}{dt^n} e^{at}= a^n e^{at}; \, n\in \mathbb N.$$
I want to know if the result is true if $n$ is a real number, i.e., $n\in \mathbb R$ ?
 A: It's one possible definition. You know the result holds true for any non-negative integer $n$, so you define it to be $a^ne^{at}$ for fractional $n$. As others have pointed out this is one of the starting points of fractional calculus. 
A: One common definition of the fractional derivative is $D^\alpha$ of $f(x) = \frac{1}{\Gamma(m-\alpha)} \frac{d}{dx} \int_0^x \frac{f(t)}{(x-t)^\alpha} dt$  (for any $0<\alpha<1)$ 
Substituting in $\alpha = n$ and $f(x) = e^{ax}$ we get
$$\frac{1}{\Gamma(1-n)} \frac{d}{dx} \int_0^x \frac{e^{ax}}{(x-t)^n} dt$$
$$=\frac{e^{ax}x^{-n}(n-ax-1)}{(n-1)\Gamma(1-n)}$$
If we set $n=\frac{1}{2}$ we get the half-derivative as
$$=\frac{2e^{ax}x^{-1/2}(ax+\frac{1}{2})}{\sqrt{\pi}}$$
While this may be strange, it may be somewhat comforting to note that the limit as $n \to 1$ in the above formula yields the answer $ae^{ax}$ as expected. Now, this isn't the only the only fractional derivative out there, and you could very well choose to simply define your answer as correct!
Note:
In general, the derivative formula  I am using is as follows: for any $m-1<\alpha<m$ where $m \in \mathbb{N}$, $\frac{1}{\Gamma(m-\alpha)} \frac{d}{dx} \int_0^x \frac{f(t)}{(x-t)^{\alpha+1-m}} dt$
This is sometimes known as the Left-Hand Definition, and a derivation can be found here 
Note 2: I have provided a comparison of the two derivatives, your's in orange and mine in red. The function $ae^{ax}$ is provided in blue to show the convergence of the the two derivatives as $n \to 1$:

