Power Rule for Derivatives

Theorem: If $$f(x)=x^n$$, $$\forall x\in R$$ and $$\forall n \in N$$, then $$\frac{d}{dx}f(x) = nx^{n-1}$$

Proof
Let $$f(x)=x^n$$,$$x\in R$$ and $$n\in N$$.

By the definition of the derivative, $$\frac{d}{dx}f(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} = \lim_{h\rightarrow 0} \frac{(x+h)^n-x^n}{h}$$.

Using the binomial theorem this simplifies to:

$$\lim_{h\rightarrow 0}\frac{({n\choose 0}x^n+{n\choose 1}x^{n-1}h+{n\choose 2}x^{n-2}h^2+\dots+{n\choose n-1}xh^{n-1}+{n\choose n}h^n)-x^n}{h}$$

$$=\lim_{h\rightarrow 0}\frac{{n\choose 1}x^{n-1}h+{n\choose 2}x^{n-2}h^2+\dots+{n\choose n-1}xh^{n-1}+{n\choose n}h^n}{h}$$

$$=\lim_{h\rightarrow 0} {n\choose 1}x^{n-1}+{n\choose 2}x^{n-2}h^1+\dots+{n\choose n-1}xh^{n-2}+{n\choose n}h^{n-1}$$

Evaluating the limit gives,

$$={n\choose 1}x^{n-1} = nx^{n-1}$$

Q.E.D.

Extension for all real number
We're gonna prove that $$f^{\,'}(x)=nx^{n-1}$$ holds for all $$n$$ real number. Actually, one has to compute again the limit $$\underset{h\to 0}{\mathop{\lim }}\,\frac{(x+h)^{n}-x^{n}}{h}.$$ Now, let's do some of algebra:

$$\frac{(x+h)^{n}-x^{n}}{h}=\frac{x^{n}}{h}\left\{ \left( 1+\frac{h}{x} \right)^{n}-1 \right\}=\frac{x^{n}}{h}\left\{ \exp \left[ n\ln \left( 1+\frac{h}{x} \right) \right]-1 \right\},$$

hence, $$\frac{(x+h)^{n}-x^{n}}{h}=x^{n}\cdot \frac{\exp \left[ n\ln \left( 1+\frac{h}{x} \right) \right]-1}{n\ln \left( 1+\frac{h}{x} \right)}\cdot \frac{n\ln \left( 1+\frac{h}{x} \right)}{\frac{h}{x}}\cdot \frac{1}{x}\to nx^{n-1}$$ as $$h\to0 .$$

Q.E.D.

(Note that we used the following facts: $$\underset{x\to 0}{\mathop{\lim }}\,\frac{e^{x}-1}{x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\ln (1+x)}{x}=1.$$ Following up on this, the above limits are easily calculated by just making the proper change of variables.)