an "informal proof"
[(x+h)n - xn]/h =
(xn + nxn-1h + (other stuff with h2 terms) - xn)/h =
(nxn-1h)/h + h2(...stuff...)/h =
nxn-1 + h(...stuff...).
no matter what the "...stuff..." is, it is clearly not infinite at any particular x, so we have:
Plato's post makes this argument rigorous, but this gives you "the general idea".
or, you can use induction:
base case: n = 1
assume that for n = k-1:
, by our induction hypothesis, and the base case,
, and we are done.
No, it isn't.
Another proof, for n an integer, would be inductive: if n= 0, then is a constant so its derivative is . Assume that, for some k, the derivative of is . Then we can differentiate by the product rule: the derivative is .
For k not an integer, we can use "logarithmic differentiation". If , then . Now, we need the fact that the derivative of ln(x) is 1/x (which why we typically start with integer powers of x). The derivative of the left side is and the derivative of the right side is . Since , we have . From that, we have .