Elegant generalization of power rule via limits Today I was thinking about how to generalize the derivative of $f(x) = x^n$. Of course, we know the following:
$$\frac{d}{dx}x^n = n\cdot x^{n-1}$$
But how can we show that the power rule behaves this way? I took to the limit process and came up with this:
$$\frac{d}{dx}x^n = \lim_{h\to 0}\frac{(x+h)^n-x^n}{h}$$
First I thought to express $(x+h)^n$ as a sum through the binomial theorem and rewrite the exression.
$$\frac{d}{dx}x^n = \lim_{h\to 0}\frac{1}{h}\left[\sum_{i=0}^n\binom{n}{i}x^nh^i\;-x^n\right]$$
Here's where I struggle to elegantly finish the generalization.
$$\frac{d}{dx}x^n = \lim_{h\to0}\frac{1}{h}\left[\binom{n}{0}x^n+\binom{n}{1}x^{n-1}h+\binom{n}{2}x^{n-2}h^2\;...\;-x^n\right]$$
Simplifying the first term:
$$\frac{d}{dx}x^n = \lim_{h\to0}\frac{1}{h}\left[{x^n}+\binom{n}{1}x^{n-1}h+\binom{n}{2}x^{n-2}h^2\;...\;-x^n\right]$$
Cancelling end terms:
$$\frac{d}{dx}x^n = \lim_{h\to0}\frac{1}{h}\left[\binom{n}{1}x^{n-1}h+\binom{n}{2}x^{n-2}h^2\;...\right]$$
Factoring out an $h$:
$$\frac{d}{dx}x^n = \lim_{h\to0}\;\frac{h}{h}\cdot\left[\binom{n}{1}x^{n-1}+\binom{n}{2}x^{n-2}h\;...\right]$$
Simplifying and using direct substitution:
$$\frac{d}{dx}x^n = \binom{n}{1}x^{n-1}+0+0\;...$$
We know this:
$$\binom{n}{1} = \frac{n!}{(n-1)!1!} = n$$
Thus:
$$\frac{d}{dx}x^n = n\cdot x^{n-1}$$
I know it may seem petty, but I hate having the points of ellipsis in a formal proof and overall it's a bit messy. How can I clean this up a bit so that it's more presentable?

If it helps, you could retain the summation notation, assuming $n\ge2$:
$$\eqalign{
  \Bigl(\sum_{i=0}^n\binom ni x^{n-i}h^i\Bigr)-x^n
  &=x^n+nx^{n-1}h+\Bigl(\sum_{i=2}^n\binom ni x^{n-i}h^i\Bigr)-x^n\cr
  &=nx^{n-1}h+h^2\sum_{i=2}^n\binom ni x^{n-i}h^{i-2}\cr}$$
and keep going from there.
