Elementary proof of Power Rule For differentiation $f'(x)=rx^{r-1}$ for $f(x)=x^r$ I could  not find any elementary proof of the Power Rule for differentiation:
Given $x,r \in R$ ,$x>0$ and a function $f(x)=x^r$, then its derivative is $f'(x)=rx^{r-1}$
Defintion:Let a sequence of rational numbers {$r_n$} tend to $r$ then $x^r=\lim_{r_n \to r}x^{r_n}$
All the proofs I had seen utilized the derivative of $logx$ but I dont think this is  elementary because one has to show that for $e=\text{Euler's number}$ , $t \in R$and $n \in N$ $\lim_{n \to \infty}(1+\frac{1}{n})^n=\lim_{t \to 0}(1+t)^\frac{1}{t}=e$ which can be proved using the Power Rule
Ofcourse the proof for the Power Rule is straight forward if one is familiar with the result of Power Rule if $r$ is a rational number and theorem  regarding  derivative of a sequence of derivatives that are uniformly convergent.

For $r\in\mathbb{R}$, to prove that $f(x)=x^{r}\implies f'(x)=rx^{r-1}$ We need two lemmas :

Lemma $(1)$ : Consider the function $E(x)=\displaystyle\sum_{n=0}^{+\infty}\frac{x^{n}}{n!}$. We claim that $E'(x)=E(x)$.

Proof :
Let $x\in\mathbb{R}$ with $h\neq0$, we define $\tau(h)=\displaystyle\frac{f(x+h)-f(x)}{h}=\frac{E(x+h)-E(x)}{h}$. Since $E(x+y)=E(x)E(y)$, then we have that :
$$
\tau(h)=\displaystyle\frac{f(x+h)-f(x)}{h}=\frac{E(x+h)-E(x)}{h}=E(x)\cdot\frac{E(h)-1}{h}
$$
Thus we need to prove that $\displaystyle\lim_{h\to 0}\frac{E(h)-1}{h}=1$. Note that :
\begin{align*}
\frac{E(h)-1}{h}&=\frac{1}{h}\left(\left(1+h+\frac{h^{2}}{2!}+\cdots\right)-1\right) \\
&=1+\frac{h}{2!}+\frac{h^{2}}{3!}+\frac{h^{3}}{4!}+\cdots \\
\implies\left|\frac{E(h)-1}{h}-1\right|&=\left|\frac{h}{2!}+\frac{h^{2}}{3!}+\frac{h^{3}}{4!}+\cdots\right| \\
&\leq\frac{|h|}{2!}+\frac{|h|^{2}}{3!}+\frac{|h|^{3}}{4!}+\cdots \\
&=\underbrace{|h|}_{\to0}\underbrace{\left(\frac{1}{2!}+\frac{|h|}{3!}+\frac{|h|^{2}}{4!}+\cdots\right)}_{\text{bounded}} 
\end{align*}
Since $\displaystyle\lim_{h\to 0}|h|=0$, and if we can show that the other part is bounded, then the overall limit is $0$. Let $0<|h|<1$, we note that the chosen interval of $h$ is due to our interest of $h$ being close to $0$. We have that :
\begin{align*}
&\left|\frac{E(h)-1}{h}-1\right|=|h|\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots\right)=|h|\underbrace{\sum_{n=2}^{+\infty}\frac{1}{n!}}_{\text{converges}} \\
\implies\lim_{h\to 0}&\left|\frac{E(h)-1}{h}-1\right|=\lim_{h\to 0}|h|\sum_{n=2}^{+\infty}\frac{1}{n!}=0
\end{align*}
Therefore, by sandwich theorem :
$$
\lim_{h\to 0}\left|\frac{E(h)-1}{h}-1\right|=0\implies\lim{h\to 0}\frac{E(h)-1}{h}=1
$$
Hence $\displaystyle\lim_{h\to 0}\tau(h)=\lim_{h\to 0}E(x)\cdot\frac{E(h)-1}{h}=E(x)$, $\forall x\in\mathbb{R}$ with $E'(x)=E(x)$.

Lemma $(2)$ : Let $g(x)=\ln(x)$, we claim that $\displaystyle g'(x)=\frac{1}{x}$, $\forall x>0$.

Proof : Let $x_{0}\in(0,+\infty)$ with $x_{0}\neq x$, let $\displaystyle\phi(x):=\frac{\ln(x)-\ln(x_{0})}{x-x_{0}}$. Moreover, there exists $t_{x}\in\mathbb{R}$ with $e^{t_{x}}=x$ and there exists $t_{0}\in\mathbb{R}$ with $e^{t_{0}}=x_{0}$. Thus, we have that :$$\displaystyle\phi(x)=\frac{t_{x}-t_{0}}{e^{t_{x}}-e^{t_{0}}}=\displaystyle\frac{1}{\displaystyle\frac{e^{t_{x}}-e^{t_{0}}}{t_{x}-t_{0}}}$$
Notice that $\ln(x)$ is continuous and that $\ln(x)=t_{x}$ and $\ln(x_{0})=t_{0}$. Thus, we have that :
\begin{align*}
&\lim_{x\to x_{0}}\ln(x)=\ln(x_{0})=t_{0} \\
\implies&\lim_{x\to x_{0}}\phi(x)=\lim_{x\to x_{0}}\frac{\ln(x)-\ln(x_{0})}{x-x_{0}}\\
&\;\qquad\qquad=\frac{1}{(e^{t})'|_{t=t_{0}}}=\frac{1}{e^{t_{0}}}=\frac{1}{x_{0}}
\end{align*}
Therefore, $g$ is differentiable at $x_{0}$ and $g'(x_{0})=\displaystyle\frac{1}{x_{0}}$, $\forall x_{0}\in(0,+\infty)$.
Conclusion : We have that $f(x)=x^{r}=e^{\ln(x^{r})}=e^{r\ln(x)}$. We note that $f$ is the composition of $e^{x}$ and $r\ln(x)$ that are both differentiable. Thus :
$$
f'(x)=(r\ln(x))'(e^{x})'|_{x=r\ln(x)} =\frac{r}{x}e^{r\ln(x)}=\frac{r}{x}x^{r}=rx^{r-1}
$$
