Show that $z^n$ is analytic. Hence find its derivative. Show that $z^n$ is analytic. Hence find its derivative.
We have $\displaystyle w=f(z)=z^n=(x+iy)^n=u(x,y)+iv(x,y)$
We are asked to find $\displaystyle \frac{dw}{dz}=f'(z)$
If $z^n$ is analytic, I can calculate $\displaystyle f'(z)=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}$
How do I open up $z^n$ to get u and v ? Taylor's expansion requires an independent variable.
 A: Of course, the method used depends on the definition of "analytic". user1337's excellent answer is based on the Definition: $f$ is analytic at $z$ if and only if
$$\lim_{h\to 0} \frac{f(z+h)-f(z)}{h}$$
exists. 
I think this is the simplest and most elementary definition of "analytic". Of course, there are other equivalent definitions. For example, it is natural to consider "analytic" as satisfying a certain system of PDE's, the Cauchy-Riemann equations, which can be written as 
$$\frac{\partial f}{\partial \overline{z}} = 0.$$
We should define the notation:
$$\frac{\partial f}{\partial \overline{z}} = \frac{1}{2}\left(\frac{\partial f}{\partial x} - \frac{1}{i}\frac{\partial f}{\partial y}\right).$$
I think it is worth understanding this point of view well but only once you have understood the point of view expressed in user1337's answer. I give the following exercise as practice.
Exercise 1: 
(a) Show that $f$ is analytic at $z$ (i.e., the limit at the beginning of my answer exists) if and only if $\frac{\partial f}{\partial \overline{z}}=0$ at $z$.
(b) Show that $\frac{\partial f}{\partial \overline{z}} = 0$ if and only if the differential $df$ (of the map $f:\mathbb{C}\to \mathbb{C}$ viewed as a map $\mathbb{R}^2\to\mathbb{R}^2$ in the natural manner) is a scalar multiple of the differential $dz$. In this case, prove that $f'(z)$ is the pertinent scalar (where $f'(z)$ is the value of the limit at the beginning of my answer).
If you have understood this point of view, then note then it remains to show that $\frac{\partial z^n}{\partial \overline{z}}=0$ in order to establish the analyticity of $z\to z^n$; furthermore, the derivative of $z\to z^n$ will be $\frac{\partial z^n}{\partial z} = \frac{1}{2}\left(\frac{\partial z^n}{\partial x} + \frac{1}{i} \frac{\partial z^n}{\partial y}\right)$ (why?). However, all of this is equivalent to evaluating the limit in user1337's answer.
Hope this helps!
A: Hint:
Consider the quotient
$$\frac{f(z_0+\Delta z)-f(z_0)}{\Delta z}=\frac{(z_0+\Delta z)^n-z_0^n}{\Delta z} $$
and apply Newton's Binomial formula.
