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If $f\,(x) = x^n$, for some positive integer $n$, then $f\,'(x) = n x^{n-1}$.
Proof:
Suppose $f(x)=c$, for some constant $c$. Then the derivative of $f(x)$ can be found as follows
$$\begin{array}{rcl} f\,'(x) & = & \displaystyle{\lim_{h \rightarrow 0} \ \frac{f\,(x+h) - f\,(x)}{h}}\\\\ & = & \displaystyle{\lim_{h \rightarrow 0} \ \frac{(x+h)^n - x^n}{h}}\\\\ \end{array}$$
Since $n$ is a positive integer, we can apply the Binomial Theorem to expand $(x+h)^n$ to obtain the following
Recall $\displaystyle{{}_nC_k = \frac{n!}{k!(n-k)!}}$ gives the number of ways to choose $k$ things from a group of $n$ things, where order doesn't matter.
$$\begin{array}{rcl} f\,'(x) & = & \displaystyle{\lim_{h \rightarrow 0} \ \frac{(x^n + {}_nC_1 x^{n-1}h + {}_nC_2 x^{n-2}h^2 + \cdots + {}_nC_{n-1} xh^{n-1} + h^n) - x^n}{h}}\\\\ & = & \displaystyle{\lim_{h \rightarrow 0} \ \frac{{}_nC_1 x^{n-1}h + {}_nC_2 x^{n-2}h^2 + \cdots + {}_nC_{n-1} xh^{n-1} + h^n}{h}}\\\\ & = & \displaystyle{\lim_{h \rightarrow 0} \ \frac{h({}_nC_1 x^{n-1} + {}_nC_2 x^{n-2}h + \cdots + {}_nC_{n-1} xh^{n-2} + h^{n-1})}{h}}\\\\ & = & \displaystyle{\lim_{h \rightarrow 0} \ \left[{}_nC_1 x^{n-1} + {}_nC_2 x^{n-2}h + \cdots + {}_nC_{n-1} xh^{n-2} + h^{n-1} \right]} \end{array}$$ With the $h$ in the denominator gone, this last limit can be evaluated by simple substitution, yielding $$\displaystyle{f\,'(x) = {}_nC_1 x^{n-1} = n x^{n-1}}$$