How can I prove $\frac {d}{dx} {x^n} = n x^{n-1}$ for $ n \in \Bbb R$ without circular reasoning? I just cannot prove that $$\frac {d}{dx} {x^n} = n x^{n-1}$$ for $ n  \in \Bbb R$.
For $n \in \Bbb{N}$, I can use the definition of a derivative :
$$\frac {d}{dx}x^n = \lim_{h \rightarrow 0} \frac{(x+h)^n - x^n}{h}$$
Now applying "Binomial Expansion" for $\displaystyle (x+h)^n=\sum_{i=0}^{n}{n \choose i }x^{n-i}h^i$ and expanding, the $x^n$ term in the numerator cancels out and the $h$ from denominator divides the entire remaining expression . Taking limit $h$ tending to $0$ gives the required result.
I have been taught that the derivative result holds for all real $n$. But I am not aware of any "formula" which can allow me to expand a binomial expression with real index. I do know about the Taylor Expansion, but if I remember correctly, it utilises the very derivative that I am trying to find.
How can I proceed ? 
 A: As stated in the comments we can use the fact that 
$$x^n=e^{n\ln{x}}$$
For all $x \in \mathbb{C}$ except $0$, $n \in \mathbb{C}$. Then, by using the chain rule, we have
$$\frac{d}{dx}\Big(e^{f(x)}\Big)=f'(x)\cdot e^{f(x)}$$
So,
$$\frac{d}{dx}\Big(x^n\Big)=\frac{d}{dx}\Big(e^{n\ln{x}}\Big)=\frac{n}{x}\cdot e^{n\ln{x}}=\frac{n}{x} \cdot x^n = n \cdot x^{n-1}$$
A: For $\boldsymbol{n\ge1}$
Bernoulli's Inequality, which is proven for integer exponents in this answer and extended to rational exponents in this answer using induction, says that for $n\ge1$,
$$
1+nx\le(1+x)^n\tag1
$$
From $(1)$, we get
$$
1+x\le\lim_{n\to\infty}\left(1+\frac xn\right)^n=e^x\tag2
$$
Therefore,
$$
x^n\left(1+\frac{nh}x\right)\le\overbrace{x^n\left(1+\frac hx\right)^n}^{(x+h)^n}\le x^ne^{nh/x}\tag3
$$
where the left inequality is $(1)$ and the right inequality is the $n^\text{th}$ power of $(2)$.
Subtracting $x^n$ and dividing by $h$ gives
$$
x^n\frac nx\stackrel{h\gtrless0}\lesseqgtr\frac{(x+h)^n-x^n}{h}\stackrel{h\gtrless0}\lesseqgtr x^n\frac nx\frac{e^{nh/x}-1}{nh/x}\tag4
$$
Applying $(9)$ gives
$$
x^n\frac nx\stackrel{h\gtrless0}\lesseqgtr\frac{(x+h)^n-x^n}{h}\stackrel{h\gtrless0}\lesseqgtr x^n\frac nx\frac1{1-nh/x}\tag5
$$
Then the Squeeze Theorem yields
$$
\bbox[5px,border:2px solid #C0A000]{\frac{\mathrm{d}}{\mathrm{d}x}x^n=nx^{n-1}}\tag6
$$

Extending to $\boldsymbol{n\lt1}$
For smaller $n$, we can use the product rule and induction. That is, suppose we know that $(6)$ holds for some $n$, then
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}x^{n-1}
&=\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac1xx^n\right)\\
&=\frac1xnx^{n-1}-\frac1{x^2}x^n\\
&=(n-1)x^{n-2}\tag7
\end{align}
$$
thus, $(6)$ holds for $n-1$.

Bounds on $\boldsymbol{\frac{e^x-1}x}$
Taking $(2)$, substituting $x\mapsto-x$, and taking reciprocals yields that for $x\lt1$,
$$
e^x\le\frac1{1-x}\tag8
$$
Combining $(2)$ and $(8)$, subtracting $1$ and dividing by $x$ gives
$$
1\stackrel{x\gtrless0}\lesseqgtr\frac{e^x-1}x\stackrel{x\gtrless0}\lesseqgtr\frac{1}{1-x}\tag9
$$
A: The next step is to prove the result for exponents of the form $1/n \text{ with } n \in \Bbb N \text{ and } n \gt 0$.  You can do this via implicit differentiation:  If $y = x^{1/n}, \text{ then } y^n = x$.  Then use the quotient rule to get the result for negative exponents and the chain rule and the two previous results (using $x^{p/q} = {(x^{1/q})}^p$) to obtain the result for all rational exponents.
Obtaining the result for irrational exponents first requires you to define exponentiation for irrational exponents.  You would get a correct definition by taking limits of rational exponents, but the definition that's much easier to work with is $x^n=e^{n \cdot\ln x}$, which you can differentiate using the chain rule and the fact (derived from the definition of $e^x$) that $d(e^x)/dx=e^x$.
And by the way, good job realizing that using a Taylor series would be circular reasoning.
