the derivative of $n^x, n\in\mathbb{R}$ using the definition of derivative I know that the derivative of $n^x$ is $n^x\times\ln n$ so i tried to show that with the definition of derivative:$$f'\left(x\right)=\dfrac{df}{dx}\left[n^x\right]\text{ for }n\in\mathbb{R}\\{=\lim_{h\rightarrow0}\dfrac{f\left(x+h\right)-f\left(x\right)}{h}}{=\lim_{h\rightarrow0}\frac{n^{x+h}-n^x}{h}}{=\lim_{h\rightarrow0}\frac{n^x\left(n^h-1\right)}{h}}{=n^x\lim_{h\rightarrow0}\frac{n^h-1}{h}}$$ now I can calculate the limit, lets:$$g\left(h\right)=\frac{n^h-1}{h}$$ $$g\left(0\right)=\frac{n^0-1}{0}=\frac{0}{0}$$$$\therefore g(0)=\frac{\dfrac{d}{dh}\left[n^h-1\right]}{\dfrac{d}{dh}\left[h\right]}=\frac{\dfrac{df\left(0\right)}{dh}\left[n^h\right]}{1}=\dfrac{df\left(0\right)}{dh}\left[n^h\right]$$
so in the end i get: $$\dfrac{df}{dx}\left[n^x\right]=n^x\dfrac{df\left(0\right)}{dx}\left[n^x\right]$$
so my question is how can i prove that $$\dfrac{df\left(0\right)}{dx}\left[n^x\right]=\ln n$$
edit:
i got 2 answers that show that using the fact that $\lim_{z \rightarrow 0}\dfrac{e^z-1}{z}=1$, so how can i prove that using the other definitions of e, i know it is definition but how can i show that this e is equal to the e of $\sum_{n=0}^\infty \frac{1}{n!}$?

If you don't have a definition of the logarithm handy (or suitable properties taken for granted), you cannot obtain the stated result because the logarithm will not appear by magic from the computation.
Assume that the formula $n^x=e^{x \log n}$ is not allowed. Then to define the powers, you can work via rationals
$$n^{p/q}=\sqrt[q]{n^p}$$ and extend to the reals by continuity.
Using this apporach, you obtain
$$\lim_{h\to0}\frac{n^h-1}h=\lim_{m\to\infty}m(\sqrt[m]n-1)$$
and you can take this as a definition of the logarithm.
$$\log n:=\lim_{m\to\infty}m(\sqrt[m]n-1).$$
