Sometimes the fact that that limit is $1$ is taken to be the definition of $e$. (That's what's done in Stewart's textbook.)
Supposing you had $\dfrac{d}{dx} 4^x$ instead of $\dfrac{d}{dx} e^x$. You'd get $4^x\cdot\lim\limits_{h\to0}\dfrac{4^h - 1}{h}$.
But the function $y=4^x$ gets steeper as you go from left to right, and the slope of its secant line through the points where $x=-1/2$ and $x=0$ is $1$; therefore the slope of the curve at $x=0$ is more than $1$. If you had $y=2^x$, you'd consider the secant line at $x=0$ and $x=1$ and conclude that the slope at $0$ is less than $1$.
So $4$ is too big, and $2$ is too small, to be the base of the natural exponential function. Somewhere between $2$ and $4$ is the right number. If $e$ is that right number, then of course the slope at $0$ is $1$.
(What this omits is how we know just where that number is, i.e. that it's $2.71828\ldots$, beyond the fact that it's between $2$ and $4$.)