Why the differentiation of $e^x$ is $e^x?$

$$\frac{d}{dx} e^x=e^x$$ Please explain simply as I haven't studied the first principle of differentiation yet, but I know the basics of differentiation.

Have a look at the series representation of $e^x$ which is $$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\dots$$ Taking derivative of this gives $$\left(e^x\right)'=\left(\sum_{n=0}^{\infty}\frac{x^n}{n!}\right)'=\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\dots\right)'$$ $$=1'+x'+\left(\frac{x^2}{2}\right)'+\left(\frac{x^3}{6}\right)'+\left(\frac{x^4}{24}\right)'+\left(\frac{x^5}{120}\right)'+\dots$$ $$\implies (e^x)'=\sum_{n=0}^{\infty}\left(\frac{x^n}{n!}\right)'$$Then, differentiating term by term gives us $$(e^x)'=0+1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\dots$$ $$=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\dots$$ $$=e^x$$ $$\implies \frac{d}{dx}e^x=e^x$$

\begin{align} \frac{d}{dx}e^x &= \frac{d}{dx}\left(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots \right)\\ &=0+\frac{1}{1!}+\frac{2x}{2!}+\frac{3x^2}{3!}+\dots\\ &=0+1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots\\ &=0 + e^x\\ &=e^x \end{align}

$$e^x=\lim_{n\to\infty}\left(1+\frac{x}{n} \right)^n$$

\begin{align} \frac{d}{dx}e^x &= \frac{d}{dx}\lim_{n\to\infty}\left(1+\frac{x}{n} \right)^n\\ &=\lim_{n\to\infty}\frac{d}{dx}\left(1+\frac{x}{n} \right)^n\\ &=\lim_{n\to\infty}n\left(1+\frac{x}{n} \right)^{n-1}\cdot\frac{1}{n}\\ &=\lim_{n\to\infty}\left(1+\frac{x}{n} \right)^{n-1}\\ &=\lim_{n\to\infty}\left(1+\frac{x}{n} \right)^{-1}\cdot\lim_{n\to\infty}\left(1+\frac{x}{n} \right)^n\\ &=1\cdot e^x\\ &=e^x \end{align}

$$\frac{d}{dx}e^x=\lim_{h\to 0}\frac{e^{x+h}-e^x}{h}=\lim_{h\to 0}e^x\left(\frac{e^h-1}{h}\right)=e^x\lim_{h\to 0}\frac{e^h-1}{h}=e^x\lim_{h\to 0}\frac{1+h-1+o(h)}{h}$$$$=e^x\lim_{h\to 0}\left(1+\underbrace{\frac{o(h)}{h}}_{\to 0}\right)=e^x$$