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 real exponents in general, I don't think you can avoid the exponential function and the logarithm because the definition of $x^r$ tends to be $\exp(r\log x)$.
However, for integer exponents there are a number of simple proofs. Here is a lesser-known one. Let $f:\mathbb{R}\mapsto\mathbb{R}$ be the function defined by $f(x)=x^n$ for all $x$, with $n \in \mathbb{N}$. We find that
\begin{align}
f'(a) &= \lim_{x \to a}\frac{f(x)-f(a)}{x-a} \\[6pt]
&= \lim_{x \to a}\frac{x^n-a^n}{x-a} \\[6pt]
&= \lim_{x \to a}\frac{(x-a)(x^{n-1}+x^{n-2}a+x^{n-3}a^2+\ldots+xa^{n-2}+a^{n-1})}{x-a} \\[6pt]
&= \lim_{x \to a}x^{n-1}+x^{n-2}a+x^{n-3}a^2+\ldots+xa^{n-2}+a^{n-1} \\[6pt]
&= a^{n-1}+a^{n-2}a+a^{n-3}a^2+\ldots+aa^{n-2}+a^{n-1}\\[6pt]
&= \underbrace{a^{n-1}+a^{n-1}+a^{n-1}+\ldots+a^{n-1}+a^{n-1}}_{\text{$n$ terms}}\\[6pt]
&= na^{n-1}\blacksquare
\end{align}
This proof can be extended to negative integer exponents by writing $x^{-n}$ as $1/x^n$ and using the quotient rule.
