how do you prove the power rule with a negative integer?

Given f(x) = x^(-n) for some positive integer n:

f '(x) = lim(h-->0) [1/(x+h)^n - 1/x^n] / h
........= lim(h-->0) {[x^n - (x+h)^n] / ([(x+h)^n * x^n]} / h, by using a common denominator for the top
........= lim(h-->0) [x^n - (x+h)^n] / [h x^n (x+h)^n]

Now, this is done as for positive exponents; use the binomial theorem
f '(x) = lim(h-->0) [x^n - (x^n + nx^(n-1) * h + {terms with higher powers of h})] / [h x^n (x+h)^n]
........= lim(h-->0) [-nx^(n-1) * h - {terms with higher powers of h})] / [h x^n (x+h)^n]
........= lim(h-->0) [-nx^(n-1) - {terms with h}] / [x^n (x+h)^n]
........= [-nx^(n-1) - {0}] / [x^n (x+0)^n]
........= -n x^(n-1) / x^(2n)
........= -n x^(-n-1), as required.

I hope this helps!

You can also get away with a little less work if you assume you know that (1/x)' = -1/x^2.
Let n be a positive integer.
Then x^(-n) = (1/x)^n, and the chain rule gives (x^{-n})' = n (1/x)^{n-1} (-1/x^2) = -n (1/x)^{n+1} = -nx^{-n-1}, where we use the chain rule and the power rule for positive integers.