Limit involving exponentials Being bored, I recently started trying to prove the exponential derivative formula by difference quotient:

$\dfrac{d}{dx}n^x=\lim\limits_{\Delta x \to 0}\dfrac{n^{x+\Delta x}-n^x}{\Delta x} = n^x\log n$

Simple algebraic manipulation (exponent rule and factoring) brought me from the difference quotient to:

$\lim\limits_{\Delta x \to 0}n^x\dfrac{n^{\Delta x}-1}{\Delta x}$

Limit of a product:

$\Bigg(\lim\limits_{\Delta x \to 0}n^x\Bigg)\Bigg(\lim\limits_{\Delta x \to 0}\dfrac{n^{\Delta x}-1}{\Delta x}\Bigg)$

And finally limit of a constant.

$n^x\Bigg(\lim\limits_{\Delta x \to 0}\dfrac{n^{\Delta x}-1}{\Delta x}\Bigg)$

This limit is where I got stuck, however. Clearly it equals $\log n$ by the well-known formula, but how can the limit be evaluated? Apologies if this is somewhat basic.
Write $n^{\Delta x}=e^{\Delta x\cdot \log n}$. Then by the definition of $\exp$ by its Taylor development:

$$\frac{e^{\Delta x\log n}-1}{\Delta x}=\frac{(\Delta x\log n)+\frac{(\Delta x\log n)^2}{2}+\cdots}{\Delta x}=\log n+\Delta x\cdot \left(\frac{(\log n)^2}{2}+\cdots\right)\overset{\Delta x\rightarrow 0}{\longrightarrow} \log n$$

Alternatively one can use L'Hôpital's rule. But that makes use of knowing the derivative of $e^x$.
We need to show that there exists some value $e$ such that
$$\displaystyle\lim_{h\to0}\frac{e^h-1}{h}=1$$
suppose
$$\displaystyle f(n)=\lim_{h\to0}\frac{n^h-1}{h}$$
then consider $f(1)$ vs. $f(100)$. Clearly $f(1)=0.$
I'm assuming that the exponential function is an increasing function, and if I can't make that assumption, then I'll define it as such (with $e\gt1$). I would like to find $n$ such that $dn^x/dx=1$ to that end I would like to evaluate (and yes, I'm also assuming that $n^x$ is a convex function)
$$\frac{n^{x}-n^{x-\Delta x}}{\Delta x}=n^x\frac{1-n^{-\Delta x}}{\Delta x}=n^x\frac{1}{n^{\Delta x}}\cdot\frac{n^{\Delta x}-1}{\Delta x}\lt n^xf(n)$$
Evaluating for $n=100$ and $\Delta x=1/2$ we have
$$\frac{1}{100^{1/2}}\cdot\frac{100^{1/2}-1}{1/2}=\frac{18}{10}\lt f(100)$$
So if $f$ is continuous then by the intermediate value theorem there exists some $e$ such that
$$0=f(1)\lt 1=f(e)\lt 1.8\lt f(100)$$
$$\therefore \lim_{\Delta x\to0}\frac{n^{\Delta x}-1}{\Delta x}=\lim_{\Delta x\to0}\frac{{(e^{\ln n})}^{\Delta x}-1}{\Delta x}=\lim_{\Delta x\to0}\frac{{e^{\ln n\cdot\Delta x}}-1}{\Delta x}$$
$$=\lim_{\Delta x\to0}\frac{{e^{\ln n\cdot\Delta x}}-1}{\ln n\cdot\Delta x}\cdot\ln n=\lim_{\ln n\cdot\Delta x\to0}\frac{{e^{\ln n\cdot\Delta x}}-1}{\ln n\cdot\Delta x}\cdot\ln n$$
$$=f(e)\cdot\ln n$$
$$=\ln n$$