New here, and so this might be a foolish question. I have observed the exponential rule, and that is that the derivative of $n^x = n^x \cdot \log(n)$.
I'd like to know why this is true, and why my proof of the derivative of $\operatorname{e}^x = \operatorname{e}^x \cdot\log(\operatorname{e})$ is wrong.
Your proof not really wrong, per se, since $$\frac{d}{dx}e^x = e^x\cdot \log(e)$$ is in fact true, as long as you realise that $\log$ is the natural logarithm, which means that $\log(e)=1$.
If you use a logarithm in any other base, say $a\neq e$, then the equality $$\frac{d}{dx} n^x = n^x\cdot \log_a(n)$$ is no longer true, and then, your proof is indeed wrong.