I have been trying to calculate the fractional derivative of $e^{ax}$ using the Liouville Left-Sided derivative, which states that, for $x>0$ and $0<n<1$, $D^n f(x) = \frac{1}{1-n} \frac{d}{dx} \int_0^x \frac{f(t)}{(x-t)^n}dt$
However, I have been quite unsuccessful. Using the standard Liouville Derivative (i.e. $D^n f(x) = \frac{1}{1-n} \frac{d}{dx} \int_{-\infty}^x \frac{f(t)}{(x-t)^n}dt$ for all real $x$ and $0<n<1$) Mathematica gets the answer $D^n e^{ax} = a^n e^{ax}$ as expected. However, when I run the same Left-Handed derivative through Mathematica I get the following
$$\frac{e^{ax}x^{-n}}{\Gamma(1-n)}\left[ax\operatorname{E}_{n-1}(ax) + (n-ax-1)\operatorname{E}_n(ax) + (ax)^n\Gamma(-n)\right]$$
Using the identity $\operatorname{E_n(ax)} = ax^{n-1}\Gamma(1-n,ax)$ on both of the exponential integrals gave the following result:
$$\frac{e^{ax}x^{-n}}{\Gamma(1-n)}\left[a^2x^{n-1}\Gamma(2-n,ax) + (anx^{n-1}-a^2x^n-ax^{n-1})\Gamma(1-n,ax) + (ax)^n\Gamma(1-n)\right]$$
Regardless, the above expressions do not seem to match $e^{ax}a^n$ numerically, so I fear I have either made a mistake here or the two chosen Liouville derivatives do not match up (which I doubt)
Note: that this is purely recreational - this is not for anything other than gaining experience working with the methods of fractional calculus. More specifically, I was attempting to answer the question found here graphically, but got stuck while experimenting with the Left-Handed Derivative. Also, I am using the definitions found here (with some minor variable substitution of course)
Edit: As noted in the tags, I am specifically looking for a proof verification, not alternate proof techniques (although alternate techniques are nevertheless appreciated!)
Note that
$$\frac d{dx}e^{ax}=ae^{ax}$$
We also have that, for $n\in\mathbb N$
$$\frac{d^n}{dx^n}e^{ax}=a^ne^{ax}$$
or, one could assume that this works for $n\in\mathbb R$, and put into notations,
$$D^ne^{ax}=a^ne^{ax}$$
This is the result we want to get, probably through the methods you are given.
I will say the above result should be correct via induction
$$D^{n-1}\frac d{dx}e^{ax}=aD^{n-1}e^{ax}=a^ne^{ax}=D^ne^{ax}$$
In my experience (not much) I will say that induction is the easiest way to go about this.
We also find that this formula should hold for more than $n\in\mathbb N$
$$D^kD^ne^{ax}=D^{k+n}e^{ax}$$
From that, we can show that it holds for $n\in\mathbb Q$
To show that it holds for $n\in\mathbb R$, assume that it is continuous.