(Disclaimer: I'm a high school student, and my knowledge of mathematics extends only to some elementary high school calculus. I don't know if what I'm about to do is valid mathematics.)

I noticed something really neat the other day.

Suppose we define $L$ as a "left-shift operator" that takes a function $f(x)$ and returns $f(x+1)$. Clearly, $(LLL\ldots LLLf)(x)=f(x+(\text{number of $L$s}))$, so it would seem a natural extension to denote $(L^hf)(x)=f(x+h)$.

Now, by the definition of the Taylor series, $f(x+h)=\sum\limits_{k=0}^\infty \frac{1}{k!}\frac{d^kf}{dx^k}\bigg|_{x}h^k$. Let's rewrite this as $\sum\limits_{k=0}^\infty \left(\frac{\left(h\frac{d}{dx}\right)^k}{k!}f\right)(x)$. Now, we can make an interesting observation: $\sum\limits_{k=0}^\infty \frac{\left(h\frac{d}{dx}\right)^k}{k!}$ is simply the Taylor series for $e^u$ with $u=h\frac{d}{dx}$. Let's rewrite the previous sum as $\left(e^{h\frac{d}{dx}}f\right)(x)$. This would seem to imply that $(L^hf)(x)=\left(e^{h\frac{d}{dx}}f\right)(x)$, or equivalently, $L=e^\frac{d}{dx}$. We might even say that $\frac{d}{dx}=\ln L$.

My question is, does what I just did have any mathematical meaning? Is it valid? I mean, I've done a bit of creative number-shuffling, but how does one make sense of exponentiating or taking the logarithm of an operator? What, if any, significance does a statement like $\frac{d}{dx}=\ln L$ have?