Show that the difference quotient of $1/x^n$ exists 
Let $n>0$ be a positive integer. For all $x\not=0$, prove that $f(x) = 1/x^n$ is differentiable at $x$ with $f^\prime(x) = -n/x^{n+1}$ by showing that the limit of the difference quotient exists.

I am having trouble seeing how I can manipulate the difference quotient in order to get a limit that exists
so far I have
$$f^\prime(x)= \lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h} =\lim_{h\rightarrow 0} \frac{1/(x+h)^n - 1/x^n}{h}$$
All help is much appreciated.
 A: Let $x\not=0$. Use the binomial theorem:
\begin{align}
\frac{\frac{1}{(x+h)^n}-\frac{1}{x^n}}{h} &=-\frac{(x+h)^n - x^n}{(x+h)^n x^n h}\\
&=-\frac{1}{h (x+h)^n x^n} \left(\sum_{k=0}^n {n\choose k} x^k h^{n-k} - x^n\right)\\
&=-\frac{1}{h (x+h)^n x^n} \sum_{k=0}^{n-1} {n\choose k} x^k h^{n-k}\\
&=-\frac{1}{(x+h)^n x^n} \sum_{k=0}^{n-1} {n\choose k} x^k h^{n-k-1}
\end{align}
Now we can safely let $h\rightarrow 0$. This makes the sum collapse to the summand with $k=n-1$ and we get
$$f^\prime(x)=-\frac{1}{x^{2n}} {n\choose n-1} x^{n-1}=-\frac{n}{x^{n+1}}$$
A: $$\lim_{h\to0}\frac{\frac{1}{(x+h)^n}-\frac{1}{x^n}}{h}=\lim_{h\to0}\frac{\frac{1}{x^n(1+\frac{h}{x})^n}-\frac{1}{x^n}}{h}=\lim_{h\to0}\frac{1}{x^n}\frac{\frac{1}{(1+\frac{h}{x})^n}-1}{h}=\lim_{h\to0}\frac{1}{x^{n+1}}\frac{\frac{1}{(1+\frac{h}{x})^n}-1}{\frac{h}{x}}.$$ Now let $\epsilon=\frac{h}{x}$. As $h\rightarrow0,\epsilon\rightarrow0$. Thus $$\lim_{h\to0}\frac{1}{x^{n+1}}\frac{\frac{1}{(1+\frac{h}{x})^n}-1}{\frac{h}{x}}=\frac{1}{x^{n+1}}\lim_{\epsilon\to0}\frac{\frac{1}{(1+\epsilon)^n}-1}{\epsilon}=-\frac{1}{x^{n+1}}\lim_{\epsilon\to0}\frac{1-\frac{1}{(1+\epsilon)^n}}{\epsilon}.$$ The reason I like this approach is because now, the limit expression is independent of $x$. It only depends on $n$, and all that remains is to prove that $$\lim_{\epsilon\to0}\frac{1-\frac{1}{(1+\epsilon)^n}}{\epsilon}=\lim_{\epsilon\to0}\frac{(1+\epsilon)^n-1}{\epsilon(1+\epsilon)^n}=n.$$ From the formula for the geometric progression, we know that $$(1+\epsilon)^n-1=[(1+\epsilon)-1]\sum_{m=0}^{n-1}(1+\epsilon)^m=\epsilon\sum_{m=0}^{n-1}(1+\epsilon)^m,$$ hence $$\lim_{\epsilon\to0}\frac{(1+\epsilon)^n-1}{\epsilon(1+\epsilon)^n}=\lim_{\epsilon\to0}\frac{\epsilon\sum_{m=0}^{n-1}(1+\epsilon)^m}{\epsilon(1+\epsilon)^n}=\lim_{\epsilon\to0}\frac{\sum_{m=0}^{n-1}(1+\epsilon)^m}{(1+\epsilon)^n}=\sum_{m=0}^{n-1}1=n$$
A: Collect fractions and use the binomial theorem.
